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Easy2Hard-Bench

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Furong Huang's Lab at UMD 14

AMC

AMC10

10A

N/A

Cagney can frost a cupcake every 20 seconds and Lacey can frost a cupcake every 30 seconds. Working together, how many cupcakes can they frost in 5 minutes?

25

Cagney can frost one in $20$ seconds, and Lacey can frost one in $30$ seconds. Working together, they can frost one in $\frac{20\cdot30}{20+30} = \frac{600}{50} = 12$ seconds. In $300$ seconds ($5$ minutes), they can frost $\fbox{25}$ cupcakes.

AMC10 First Half

AMC10 A

AMC

AMC10

10A

N/A

Pablo, Sofia, and Mia got some candy eggs at a party. Pablo had three times as many eggs as Sofia, and Sofia had twice as many eggs as Mia. Pablo decides to give some of his eggs to Sofia and Mia so that all three will have the same number of eggs. What fraction of his eggs should Pablo give to Sofia?

\frac{1}{6}

Assign a variable to the number of eggs Mia has, say $m$. Then, because we are given that Sofia has twice the number of eggs Mia has, Sofia has $2m$ eggs, and Pablo, having three times the number of eggs as Sofia, has $6m$ eggs. For them to all have the same number of eggs, they must each have $\frac{m+2m+6m}{3} = 3m$ eggs. This means Pablo must give $2m$ eggs to Mia and $m$ eggs to Sofia, so the answer is $\frac{m}{6m} = \fbox{\frac{1}{6}}$

AMC10 First Half

AMC10 A

AMC

AMC10

10B

N/A

Debra flips a fair coin repeatedly, keeping track of how many heads and how many tails she has seen in total, until she gets either two heads in a row or two tails in a row, at which point she stops flipping. What is the probability that she gets two heads in a row but she sees a second tail before she sees a second head？

\frac{1}{24}

We first want to find out which sequences of coin flips satisfy the given condition. For Debra to see the second tail before the second head, her first flip can't be heads, as that would mean she would either end with double tails before seeing the second head, or would see two heads before she sees two tails. Therefore, her first flip must be tails. The shortest sequence of flips by which she can get two heads in a row and see the second tail before she sees the second head is $THTHH$, which has a probability of $\frac{1}{2^5} = \frac{1}{32}$. Furthermore, she can prolong her coin flipping by adding an extra $TH$, which itself has a probability of $\frac{1}{2^2} = \frac{1}{4}$. Since she can do this indefinitely, this gives an infinite geometric series with a first term of $\frac{1}{32}$ and a common ratio of $\frac{1}{4}$, which means the answer (by the infinite geometric series sum formula) is $\frac{\frac{1}{32}}{1-\frac{1}{4}} = \fbox{\frac{1}{24}}$.

AMC10 Final Problems

AMC10 B

HMMT

HMMT-Feb

guts

Feb

Let $k$ be the answer to this problem. The probability that an integer chosen uniformly at random from $\{1,2, \ldots, k\}$ is a multiple of 11 can be written as $\frac{a}{b}$ for relatively prime positive integers $a$ and $b$. Compute $100 a+b$.

1000

Solution: We write $k=11 q+r$ for integers $q, r$ with $0 \leq r<11$. There are $q$ multiples of 11 from 1 to $k$, inclusive, so our probability is $\frac{a}{b}=\frac{q}{11 q+r}$. Let $d=\operatorname{gcd}(q, r)=\operatorname{gcd}(q, 11 q+r)$, so that the fraction $\frac{q / d}{(11 q+r) / d}$ is how we would write $\frac{q}{11 q+r}$ in simplified form. Since we require that $a$ and $b$ be relatively prime, we find $a=\frac{q}{d}$ and $b=\frac{11 q+r}{d}$. Plugging these into the equation $k=100 a+b$, we find $11 q+r=100 \frac{q}{d}+\frac{11 q+r}{d}$, or $d(11 q+r)=111 q+r$. Since $d$ divides $r$ and $r \leq 10$, we have $d \leq 10$. If we test the case $d=10$, our equation becomes $q=9 r$. Since $r=10$ is the only valid value that is a multiple of $d$, we get $q=90$ and $k=1000.10$ is, in fact, the $\operatorname{gcd}$ of $q$ and $r$, so we have found that\\ $k=1000$ satisfies the problem. Testing other values of $d$ does not produce a valid answer. $\fbox{1000}$.

HMMT Feb Guts

HMMT-Feb Guts

AMC

AMC12

12B

N/A

Let $x$ and $y$ be positive integers such that $7x^5 = 11y^{13}.$ The minimum possible value of $x$ has a prime factorization $a^cb^d.$ What is $a + b + c + d?$

31

Substitute $a^cb^d$ into $x$. We then have $7(a^{5c}b^{5d}) = 11y^{13}$. Divide both sides by $7$, and it follows that: \[(a^{5c}b^{5d}) = \frac{11y^{13}}{7}.\] Note that because $11$ and $7$ are prime, the minimum value of $x$ must involve factors of $7$ and $11$ only. Thus, we try to look for the lowest power $p$ of $11$ such that $13p + 1 \equiv 0 \pmod{5}$, so that we can take $11^{13p + 1}$ to the fifth root. Similarly, we want to look for the lowest power $n$ of $7$ such that $13n - 1 \equiv 0 \pmod{5}$. Again, this allows us to take the fifth root of $7^{13n - 1}$. Obviously, we want to add $1$ to $13p$ and subtract $1$ from $13n$ because $11^{13p}$ and $7^{13n}$ are multiplied by $11$ and divided by $7$, respectively. With these conditions satisfied, we can simply multiply $11^{p}$ and $7^{n}$ and substitute this quantity into $y$ to attain our answer. We can simply look for suitable values for $p$ and $n$. We find that the lowest $p$, in this case, would be $3$ because $13(3) + 1 \equiv 0 \pmod{5}$. Moreover, the lowest $q$ should be $2$ because $13(2) - 1 \equiv 0 \pmod{5}$. Hence, we can substitute the quantity $11^{3} \cdot 7^{2}$ into $y$. Doing so gets us: \[(a^{5c}b^{5d}) = \frac{11(11^{3} \cdot 7^{2})^{13}}{7} = 11^{40} \cdot 7^{25}.\] Taking the fifth root of both sides, we are left with $a^cb^d = 11^{8} \cdot 7^{5}$. $a + b + c + d = 11 + 7 + 8 + 5 = \fbox{31}$

AMC12 Second Half

AMC12 B

AMC

AMC8

8

N/A

Aunt Anna is $42$ years old. Caitlin is $5$ years younger than Brianna, and Brianna is half as old as Aunt Anna. How old is Caitlin?

16

If Brianna is half as old as Aunt Anna, then Brianna is $\frac{42}{2}$ years old, or $21$ years old. If Caitlin is $5$ years younger than Brianna, she is $21-5$ years old, or $16$. So, the answer is $\fbox{16}$

AMC8 First Half

AMC8

HMMT

HMMT-Nov

gen

Nov

Let $A B C D$ be a unit square. A circle with radius $\frac{32}{49}$ passes through point $D$ and is tangent to side $A B$ at point $E$. Then $D E=\frac{m}{n}$, where $m, n$ are positive integers and $\operatorname{gcd}(m, n)=1$. Find $100 m+n$.

807

Solution: Let $O$ be the center of the circle and let $F$ be the intersection of lines $O E$ and $C D$. Also let $r=32 / 49$ and $x=D F$. Then we know \[ x^{2}+(1-r)^{2}=D F^{2}+O F^{2}=D O^{2}=r^{2} \] which implies that $x^{2}+1-2 r=0$, or $1+x^{2}=2 r$. Now, \[ D E=\sqrt{D F^{2}+E F^{2}}=\sqrt{1+x^{2}}=\sqrt{2 r}=\sqrt{64 / 49}=8 / 7 \] $\fbox{807}$.

HMMT Nov Easy

HMMT-Nov General

AMC

AMC8

8

N/A

How many positive factors does $23,232$ have?

42

We can first find the prime factorization of $23,232$, which is $2^6\cdot3^1\cdot11^2$. Now, we add one to our powers and multiply. Therefore, the answer is $(6+1)\cdot(1+1)\cdot(2+1)=7\cdot2\cdot3=\fbox{42}$ Note: 23232 is a large number, so we can look for shortcuts to factor it. One way to factor it quickly is to use 3 and 11 divisibility rules to observe that $23232 = 3 \cdot 7744 = 3 \cdot 11 \cdot 704 = 3 \cdot 11^2 \cdot 64 = 3^1 \cdot 11^2 \cdot 2^6$. Another way is to spot the "32" and compute that $23232 = 32\cdot(101 + 10000/16) = 32\cdot (101+ 5^4) = 32\cdot 726 = 32 \cdot 11 \cdot 66$.

AMC8 Second Half

AMC8

HMMT

HMMT-Feb

comb

Feb

Anne-Marie has a deck of 16 cards, each with a distinct positive factor of 2002 written on it. She shuffles the deck and begins to draw cards from the deck without replacement. She stops when there exists a nonempty subset of the cards in her hand whose numbers multiply to a perfect square. What is the expected number of cards in her hand when she stops?

\frac{837}{208}

Solution: Note that $2002=2 \cdot 7 \cdot 11 \cdot 13$, so that each positive factor of 2002 is included on exactly one card. Each card can identified simply by whether or not it is divisible by each of the 4 primes, and we can uniquely achieve all of the $2^{4}$ possibilities. Also, when considering the product of the values on many cards, we only care about the values of the exponents in the prime factorization modulo 2 , as we have a perfect square exactly when each exponent is even. Now suppose Anne-Marie has already drawn $k$ cards. Then there are $2^{k}$ possible subsets of cards from those she has already drawn. Note that if any two of these subsets have products with the same four exponents modulo 2, then taking the symmetric difference yields a subset of cards in her hand where all four exponents are $0(\bmod 2)$, which would cause her to stop. Now when she draws the $(k+1)$ th card, she achieves a perfect square subset exactly when the the exponents modulo 2 match those from a subset of the cards she already has. Thus if she has already drawn $k$ cards, she will not stop if she draws one of $16-2^{k}$ cards that don't match a subset she already has. Let $p_{k}$ be the probability that Anne-Marie draws at least $k$ cards. We have the recurrence \[ p_{k+2}=\frac{16-2^{k}}{16-k} p_{k+1} \] because in order to draw $k+2$ cards, the $(k+1)$ th card, which is drawn from the remaining $16-k$ cards, must not be one of the $16-2^{k}$ cards that match a subset of Anne-Marie's first $k$ cards. We now compute \[ \begin{aligned} & p_{1}=1 \\ & p_{2}=\frac{15}{16} \\ & p_{3}=\frac{14}{15} p_{2}=\frac{7}{8} \\ & p_{4}=\frac{12}{14} p_{3}=\frac{3}{4} \\ & p_{5}=\frac{8}{13} p_{4}=\frac{6}{13} \\ & p_{6}=0 \end{aligned} \] The expected number of cards that Anne-Marie draws is \[ p_{1}+p_{2}+p_{3}+p_{4}+p_{5}=1+\frac{15}{16}+\frac{7}{8}+\frac{3}{4}+\frac{6}{13}=\frac{837}{208} \] $\fbox{\frac{837}{208}}$.

HMMT Feb Hard

HMMT-Feb Combinatorics

AMC

AMC8

8

N/A

What is the sum of the mean, median, and mode of the numbers $2,3,0,3,1,4,0,3$?

7.5

Putting the numbers in numerical order we get the list $0,0,1,2,3,3,3,4.$ The mode is $3.$ The median is $\frac{2+3}{2}=2.5.$ The average is $\frac{0+0+1+2+3+3+3+4}{8}=\frac{16}{8}=2.$ The sum of all three is $3+2.5+2=\fbox{7.5}$

AMC8 First Half

AMC8

AMC

AMC12

12A

N/A

On a sheet of paper, Isabella draws a circle of radius $2$, a circle of radius $3$, and all possible lines simultaneously tangent to both circles. Isabella notices that she has drawn exactly $k \ge 0$ lines. How many different values of $k$ are possible?

5

Isabella can get $0$ lines if the circles are concentric, $1$ if internally tangent, $2$ if overlapping, $3$ if externally tangent, and $4$ if non-overlapping and not externally tangent. There are $\fbox{5}$ values of $k$.

AMC12 Second Half

AMC12 A

AMC

AMC10

10B

N/A

Mary chose an even $4$-digit number $n$. She wrote down all the divisors of $n$ in increasing order from left to right: $1,2,\ldots,\dfrac{n}{2},n$. At some moment Mary wrote $323$ as a divisor of $n$. What is the smallest possible value of the next divisor written to the right of $323$?

340

Let $d$ be the next divisor written to the right of $323.$ If $\gcd(323,d)=1,$ then \[n\geq323d>323^2>100^2=10000,\] which contradicts the precondition that $n$ is a $4$-digit number. It follows that $\gcd(323,d)>1.$ Since $323=17\cdot19,$ the smallest possible value of $d$ is $17\cdot20=\fbox{340},$ from which \[n=\operatorname{lcm}(323,d)=17\cdot19\cdot20=6460.\]

AMC10 Final Problems

AMC10 B

AIME

AIME

II

N/A

For each integer $n\geq3$, let $f(n)$ be the number of $3$-element subsets of the vertices of the regular $n$-gon that are the vertices of an isosceles triangle (including equilateral triangles). Find the sum of all values of $n$ such that $f(n+1)=f(n)+78$.

245

Considering $n \pmod{6}$, we have the following formulas: $n\equiv 0$: $\frac{n(n-4)}{2} + \frac{n}{3}$ $n\equiv 2, 4$: $\frac{n(n-2)}{2}$ $n\equiv 3$: $\frac{n(n-3)}{2} + \frac{n}{3}$ $n\equiv 1, 5$: $\frac{n(n-1)}{2}$ To derive these formulas, we note the following: Any isosceles triangle formed by the vertices of our regular $n$-sided polygon $P$ has its sides from the set of edges and diagonals of $P$. Notably, as two sides of an isosceles triangle must be equal, it is important to use the property that same-lengthed edges and diagonals come in groups of $n$, unless $n$ is even when one set of diagonals (those which bisect the polygon) comes in a group of $\frac{n}{2}$. Three properties hold true of $f(n)$: When $n$ is odd there are $\frac{n(n-1)}{2}$ satisfactory subsets (This can be chosen with $n$ choices for the not-base vertex, and $\frac{n-1}{2}$ for the pair of equal sides as we have $n-1$ edges to choose from, and we must divide by 2 for over-count).* Another explanation: For any diagonal or side of the polygon chosen as the base of the isosceles triangle, there is exactly 1 isosceles triangle that can be formed. So, the total number of satisfactory subsets is $\dbinom{n}{2}=\dfrac{n(n-1)}{2}.$ When $n$ is even there are $\frac{n(n-2)}{2}$ satisfactory subsets (This can be chosen with $n$ choices for the not-base vertex, and $\frac{n-2}{2}$ for the pair of equal sides as we have $n-1$ edges to choose from, one of them which is not satisfactory (the bisecting edge), and we must divide by 2 for over-count). When $n$ is a multiple of three we additionally over-count equilateral triangles, of which there are $\frac{n}{3}$. As we count them three times, we are two times over, so we subtract $\frac{2n}{3}$. Considering the six possibilities $n \equiv 0,1,2,3,4,5 \pmod{6}$ and solving, we find that the only valid solutions are $n = 36, 52, 157$, from which the answer is $36 + 52 + 157 = \fbox{245}$.

Very Hard AIME Problems

AIME

AMC

AMC10

10B

N/A

Bernardo chooses a three-digit positive integer $N$ and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer $S$. For example, if $N = 749$, Bernardo writes the numbers $10,\!444$ and $3,\!245$, and LeRoy obtains the sum $S = 13,\!689$. For how many choices of $N$ are the two rightmost digits of $S$, in order, the same as those of $2N$?

25

First, we can examine the units digits of the number base 5 and base 6 and eliminate some possibilities. Say that $N \equiv a \pmod{6}$ also that $N \equiv b \pmod{5}$ Substituting these equations into the question and setting the units digits of $2N$ and $S$ equal to each other, it can be seen that $b < 5$ (because otherwise $a$ and $b$ will have different parities), and thus $a=b$. $N \equiv a \pmod{6}$, $N \equiv a \pmod{5}$, $\implies N=a \pmod{30}$, $0 \le a \le 4$ Therefore, $N$ can be written as $30x+y$ and $2N$ can be written as $60x+2y$ Just keep in mind that $y$ can be one of five choices: $0, 1, 2, 3,$ or $4$, ; Also, we have already found which digits of $y$ will add up into the units digits of $2N$. Now, examine the tens digit, $x$ by using $\mod{25}$ and $\mod{36}$ to find the tens digit (units digits can be disregarded because $y=0,1,2,3,4$ will always work) Then we take $N=30x+y$ $\mod{25}$ and $\mod{36}$ to find the last two digits in the base $5$ and $6$ representation. \[N \equiv 30x \pmod{36}\] \[N \equiv 30x \equiv 5x \pmod{25}\] Both of those must add up to \[2N\equiv60x \pmod{100}\] ($33 \ge x \ge 4$) Now, since $y=0,1,2,3,4$ will always work if $x$ works, then we can treat $x$ as a units digit instead of a tens digit in the respective bases and decrease the mods so that $x$ is now the units digit. \[N \equiv 5x \pmod{6}\] \[N \equiv 6x \equiv x \pmod{5}\] \[2N\equiv 6x \pmod{10}\] Say that $x=5m+n$ (m is between 0-6, n is 0-4 because of constraints on x) Then \[N \equiv 5m+n \pmod{5}\] \[N \equiv 25m+5n \pmod{6}\] \[2N\equiv30m + 6n \pmod{10}\] and this simplifies to \[N \equiv n \pmod{5}\] \[N \equiv m+5n \pmod{6}\] \[2N\equiv 6n \pmod{10}\] From careful inspection, this is true when $n=0, m=6$ $n=1, m=6$ $n=2, m=2$ $n=3, m=2$ $n=4, m=4$ This gives you $5$ choices for $x$, and $5$ choices for $y$, so the answer is $5* 5 = \fbox{25}$

AMC10 Final Problems

AMC10 B

AMC

AMC10

10B

N/A

The average age of $33$ fifth-graders is $11$. The average age of $55$ of their parents is $33$. What is the average age of all of these parents and fifth-graders?

24.75

The sum of the ages of the fifth graders is $33 * 11$, while the sum of the ages of the parents is $55 * 33$. Therefore, the total sum of all their ages must be $2178$, and given $33 + 55 = 88$ people in total, their average age is $\frac{2178}{88} = \frac{99}{4} = \fbox{24.75}$.

AMC10 First Half

AMC10 B

AMC

AMC8

8

N/A

Karl's car uses a gallon of gas every $35$ miles, and his gas tank holds $14$ gallons when it is full. One day, Karl started with a full tank of gas, drove $350$ miles, bought $8$ gallons of gas, and continued driving to his destination. When he arrived, his gas tank was half full. How many miles did Karl drive that day?

525

Since he uses a gallon of gas every $35$ miles, he had used $\frac{350}{35} = 10$ gallons after $350$ miles. Therefore, after the first leg of his trip he had $14 - 10 = 4$ gallons of gas left. Then, he bought $8$ gallons of gas, which brought him up to $12$ gallons of gas in his gas tank. When he arrived, he had $\frac{1}{2} \cdot 14 = 7$ gallons of gas. So he used $5$ gallons of gas on the second leg of his trip. Therefore, the second part of his trip covered $5 \cdot 35 = 175$ miles. Adding this to the $350$ miles, we see that he drove $350 + 175 = \fbox{525}$ miles.

AMC8 Second Half

AMC8

HMMT

HMMT-Nov

guts

Nov

A right triangle and a circle are drawn such that the circle is tangent to the legs of the right triangle. The circle cuts the hypotenuse into three segments of lengths 1, 24, and 3, and the segment of length 24 is a chord of the circle. Compute the area of the triangle.

192

Solution 1: Let the triangle be $\triangle A B C$, with $A C$ as the hypotenuse, and let $D, E, F, G$ be on sides $A B, B C, A C$, $A C$, respectively, such that they all lie on the circle. We have $A G=1, G F=24$, and $F C=3$. By power of a point, we have \[ \begin{aligned} & A D=\sqrt{A G \cdot A F}=\sqrt{1(1+24)}=5 \\ & C E=\sqrt{C F \cdot C G}=\sqrt{3(3+24)}=9 \end{aligned} \] Now, let $B D=B E=x$. By the Pythagorean Theorem, we get that \[ \begin{aligned} (x+5)^{2}+(x+9)^{2} & =28^{2} \\ (x+5)^{2}+(x+9)^{2}-((x+9)-(x+5))^{2} & =28^{2}-4^{2} \\ 2(x+5)(x+9) & =768 \\ (x+5)(x+9) & =384 \end{aligned} \] The area of $\triangle A B C$ is $\frac{1}{2}(x+5)(x+9)=\frac{1}{2} \cdot 384=192$. $\fbox{192}$.

HMMT Nov Guts

HMMT-Nov Guts

HMMT

HMMT-Feb

guts

Feb

Jarris is a weighted tetrahedral die with faces $F_{1}, F_{2}, F_{3}, F_{4}$. He tosses himself onto a table, so that the probability he lands on a given face is proportional to the area of that face (i.e. the probability he lands on face $F_{i}$ is $\frac{\left[F_{i}\right]}{\left[F_{1}\right]+\left[F_{2}\right]+\left[F_{3}\right]+\left[F_{4}\right]}$ where $[K]$ is the area of $\left.K\right)$. Let $k$ be the maximum distance any part of Jarris is from the table after he rolls himself. Given that Jarris has an inscribed sphere of radius 3 and circumscribed sphere of radius 10, find the minimum possible value of the expected value of $k$.

12

Solution: Since the maximum distance to the table is just the height, the expected value is equal to $\frac{\sum_{i=1}^{4} h_{i}\left[F_{i}\right]}{\sum_{i=1}^{4}\left[F_{i}\right]}$. Let $V$ be the volume of Jarris. Recall that $V=\frac{1}{3} h_{i}\left[F_{i}\right]$ for any $i$, but also $V=\frac{r}{3}\left(\sum_{i=1}^{4}\left[F_{i}\right]\right)$ where $r$ is the inradius (by decomposing into four tetrahedra with a vertex at the incenter). Therefore \[ \frac{\sum_{i=1}^{4} h_{i}\left[F_{i}\right]}{\sum_{i=1}^{4}\left[F_{i}\right]}=\frac{12 V}{3 V / r}=4 r=12 \] $\fbox{12}$.

HMMT Feb Guts

HMMT-Feb Guts

AMC

AMC10

10A

N/A

Let $T$ be the triangle in the coordinate plane with vertices $(0,0), (4,0),$ and $(0,3).$ Consider the following five isometries (rigid transformations) of the plane: rotations of $90^{\circ}, 180^{\circ},$ and $270^{\circ}$ counterclockwise around the origin, reflection across the $x$-axis, and reflection across the $y$-axis. How many of the $125$ sequences of three of these transformations (not necessarily distinct) will return $T$ to its original position? (For example, a $180^{\circ}$ rotation, followed by a reflection across the $x$-axis, followed by a reflection across the $y$-axis will return $T$ to its original position, but a $90^{\circ}$ rotation, followed by a reflection across the $x$-axis, followed by another reflection across the $x$-axis will not return $T$ to its original position.)

12

[asy] size(10cm); Label f; f.p=fontsize(6); xaxis(-6,6,Ticks(f, 2.0)); yaxis(-6,6,Ticks(f, 2.0)); filldraw(origin--(4,0)--(0,3)--cycle, gray, black+linewidth(1)); [/asy] First, any combination of motions we can make must reflect $T$ an even number of times. This is because every time we reflect $T$, it changes orientation. Once $T$ has been flipped once, no combination of rotations will put it back in place because it is the mirror image; however, flipping it again changes it back to the original orientation. Since we are only allowed $3$ transformations and an even number of them must be reflections, we either reflect $T$ $0$ times or $2$ times. Case 1: $0$ reflections on $T$. In this case, we must use $3$ rotations to return $T$ to its original position. Notice that our set of rotations, $\{90^\circ,180^\circ,270^\circ\}$, contains every multiple of $90^\circ$ except for $0^\circ$. We can start with any two rotations $a,b$ in $\{90^\circ,180^\circ,270^\circ\}$ and there must be exactly one $c \equiv -a - b \pmod{360^\circ}$ such that we can use the three rotations $(a,b,c)$ which ensures that $a + b + c \equiv 0^\circ \pmod{360^\circ}$. That way, the composition of rotations $a,b,c$ yields a full rotation. For example, if $a = b = 90^\circ$, then $c \equiv -90^\circ - 90^\circ = -180^\circ \pmod{360^\circ}$, so $c = 180^\circ$ and the rotations $(90^\circ,90^\circ,180^\circ)$ yields a full rotation. The only case in which this fails is when $c$ would have to equal $0^\circ$. This happens when $(a,b)$ is already a full rotation, namely, $(a,b) = (90^\circ,270^\circ),(180^\circ,180^\circ),$ or $(270^\circ,90^\circ)$. However, we can simply subtract these three cases from the total. Selecting $(a,b)$ from $\{90^\circ,180^\circ,270^\circ\}$ yields $3 \cdot 3 = 9$ choices, and with $3$ that fail, we are left with $6$ combinations for case $1$. Case 2: $2$ reflections on $T$. In this case, we first eliminate the possibility of having two of the same reflection. Since two reflections across the x-axis maps $T$ back to itself, inserting a rotation before, between, or after these two reflections would change $T$'s final location, meaning that any combination involving two reflections across the x-axis would not map $T$ back to itself. The same applies to two reflections across the y-axis. Therefore, we must use one reflection about the x-axis, one reflection about the y-axis, and one rotation. Since a reflection about the x-axis changes the sign of the y component, a reflection about the y-axis changes the sign of the x component, and a $180^\circ$ rotation changes both signs, these three transformation composed (in any order) will suffice. It is therefore only a question of arranging the three, giving us $3! = 6$ combinations for case 2. Combining both cases we get $6+6=\fbox{12}$.

AMC10 Final Problems

AMC10 A

AMC

AMC12

12B

N/A

What is the value in simplest form of the following expression?\[\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7}\]

10

We have \[\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7} = \sqrt{1} + \sqrt{4} + \sqrt{9} + \sqrt{16}\ = 1 + 2 + 3 + 4 = \fbox{10}.\] Note: This comes from the fact that the sum of the first $n$ odds is $n^2$.

AMC12 First Half

AMC12 B

AMC

AMC10

10A

N/A

For how many integers $n$ between $1$ and $50$, inclusive, is \[\frac{(n^2-1)!}{(n!)^n}\] an integer? (Recall that $0! = 1$.)

34

The main insight is that \[\frac{(n^2)!}{(n!)^{n+1}}\] is always an integer. This is true because it is precisely the number of ways to split up $n^2$ objects into $n$ unordered groups of size $n$. Thus, \[\frac{(n^2-1)!}{(n!)^n}=\frac{(n^2)!}{(n!)^{n+1}}\cdot\frac{n!}{n^2}\] is an integer if $n^2 \mid n!$, or in other words, if $\frac{(n-1)!}{n}$, is an integer. This condition is false precisely when $n=4$ or $n$ is prime, by Wilson's Theorem. There are $15$ primes between $1$ and $50$, inclusive, so there are $15 + 1 = 16$ terms for which \[\frac{(n^2-1)!}{(n!)^{n}}\] is potentially not an integer. It can be easily verified that the above expression is not an integer for $n=4$ as there are more factors of $2$ in the denominator than the numerator. Similarly, it can be verified that the above expression is not an integer for any prime $n=p$, as there are more factors of p in the denominator than the numerator. Thus all $16$ values of n make the expression not an integer and the answer is $50-16=\fbox{34}$. SideNote: Another method to prove that \[\frac{(n^2)!}{(n!)^{n+1}}\] is always an integer is instead as follows using Number Theory. Notice that $n$ will divide the numerator $n+1$ times, since $n^2 = n \times n$ contains not one but two factors of $n.$ Also, for $a < n,$ notice that $a$ divides $(n^2)!$ at least \[\lfloor \frac{n^2}{a} \rfloor \ge \lfloor \frac{n^2}{n-1} \rfloor \ge \lfloor \frac{n^2 - 1}{n-1} \rfloor \ge n+1\] times. Thus, each integer from $1$ to $n$ will divide $(n^2)!$ at least $n+1$ times, which proves such a lemma. $\square{}$

AMC10 Final Problems

AMC10 A

AMC

AMC10

10A

N/A

It takes Anna $30$ minutes to walk uphill $1$ km from her home to school, but it takes her only $10$ minutes to walk from school to her home along the same route. What is her average speed, in km/hr, for the round trip?

3

Since she walked $1$ km to school and $1$ km back home, her total distance is $1+1=2$ km. Since she spent $30$ minutes walking to school and $10$ minutes walking back home, her total time is $30+10=40$ minutes = $\frac{40}{60}=\frac{2}{3}$ hours. Therefore her average speed in km/hr is $\frac{2}{\frac{2}{3}}=\fbox{3}$.

AMC10 First Half

AMC10 A

HMMT

HMMT-Nov

team

Nov

In $\triangle A B C$, the external angle bisector of $\angle B A C$ intersects line $B C$ at $D . E$ is a point on ray $\overrightarrow{A C}$ such that $\angle B D E=2 \angle A D B$. If $A B=10, A C=12$, and $C E=33$, compute $\frac{D B}{D E}$.

\frac{2}{3}

Let $F$ be a point on ray $\overrightarrow{C A}$ such that $\angle A D F=\angle A D B . \triangle A D F$ and $\triangle A D B$ are congruent, so $A F=10$ and $D F=D B$. So, $C F=C A+A F=22$. Since $\angle F D C=2 \angle A D B=\angle E D C$, by the angle bisector theorem we compute $\frac{D F}{D E}=\frac{C F}{C E}=\frac{22}{33}=\frac{2}{3}$. $\fbox{\frac{2}{3}}$.

HMMT Nov Team

HMMT-Nov Team

HMMT

HMMT-Feb

geo

Feb

Let $\Omega$ and $\omega$ be circles with radii 123 and 61 , respectively, such that the center of $\Omega$ lies on $\omega$. A chord of $\Omega$ is cut by $\omega$ into three segments, whose lengths are in the ratio $1: 2: 3$ in that order. Given that this chord is not a diameter of $\Omega$, compute the length of this chord.

42

Solution: Denote the center of $\Omega$ as $O$. Let the chord intersect the circles at $W, X, Y, Z$ so that $W X=t, X Y=2 t$, and $Y Z=3 t$. Notice that $Y$ is the midpoint of $W Z$; hence $\overline{O Y} \perp \overline{W X Y Z}$. The fact that $\angle O Y X=90^{\circ}$ means $X$ is the antipode of $O$ on $\omega$, so $O X=122$. Now applying power of point to $X$ with respect to $\Omega$ gives \[ 245=123^{2}-O X^{2}=W X \cdot X Z=5 t^{2} \Longrightarrow t=7 \] Hence the answer is $6 t=42$. $\fbox{42}$.

HMMT Feb Easy

HMMT-Feb Geometry

HMMT

HMMT-Nov

guts

Nov

Trapezoid $A B C D$, with bases $A B$ and $C D$, has side lengths $A B=28, B C=13, C D=14$, and $D A=15$. Let diagonals $A C$ and $B D$ intersect at $P$, and let $E$ and $F$ be the midpoints of $A P$ and $B P$, respectively. Find the area of quadrilateral $C D E F$.

112

Note that $E F$ is a midline of triangle $A P B$, so $E F$ is parallel to $A B$ and $E F=\frac{1}{2} A B=14=C D$. We also have that $E F$ is parallel to $C D$, and so $C D E F$ is a parallelogram. From this, we have $E P=P C$ as well, so $\frac{C E}{C A}=\frac{2}{3}$. It follows that the height from $C$ to $E F$ is $\frac{2}{3}$ of the height from $C$ to $A B$. We can calculate that the height from $C$ to $A B$ is 12 , so the height from $C$ to $E F$ is 8 . Therefore $C D E F$ is a parallelogram with base 14 and height 8 , and its area is $14 \cdot 8=112$. $\fbox{112}$.

HMMT Nov Guts

HMMT-Nov Guts

AMC

AMC10

10A

N/A

A regular hexagon has side length 6. Congruent arcs with radius 3 are drawn with the center at each of the vertices, creating circular sectors as shown. The region inside the hexagon but outside the sectors is shaded as shown What is the area of the shaded region? [asy] size(125); defaultpen(linewidth(0.8)); path hexagon=(2*dir(0))--(2*dir(60))--(2*dir(120))--(2*dir(180))--(2*dir(240))--(2*dir(300))--cycle; fill(hexagon,grey); for(int i=0;i<=5;i=i+1) { path arc=2*dir(60*i)--arc(2*dir(60*i),1,120+60*i,240+60*i)--cycle; unfill(arc); draw(arc); } draw(hexagon,linewidth(1.8));[/asy]

54\sqrt{3}-18\pi

The area of the hexagon is equal to $\dfrac{3(6)^2\sqrt{3}}{2}=54\sqrt{3}$ by the formula for the area of a hexagon. We note that each interior angle of the regular hexagon is $120^\circ$ which means that each sector is $\dfrac{1}{3}$ of the circle it belongs to. The area of each sector is $\dfrac{9\pi}{3}=3\pi$. The area of all six is $6\times 3\pi=18\pi$. The shaded area is equal to the area of the hexagon minus the sum of the area of all the sectors, which is equal to $\fbox{54\sqrt{3}-18\pi}$

AMC10 Second Half

AMC10 A

HMMT

HMMT-Nov

guts

Nov

Let $P$ be a polynomial with integer coefficients such that $P(0)+P(90)=2018$. Find the least possible value for $|P(20)+P(70)|$.

782

First, note that $P(x)=x^{2}-3041$ satisfy the condition and gives $|P(70)+P(20)|=|4900+400-6082|=$ 782. To show that 782 is the minimum, we show $2800 \mid P(90)-P(70)-P(20)+P(0)$ for every $P$, since -782 is the only number in the range $[-782,782]$ that is congruent to 2018 modulo 2800 . Proof: It suffices to show that $2800 \mid 90^{n}-70^{n}-20^{n}+0^{n}$ for every $n \geq 0$ (having $0^{0}=1$ ). Let $Q(n)=90^{n}-70^{n}-20^{n}+0^{n}$, then we note that $Q(0)=Q(1)=0, Q(2)=2800$, and $Q(3)=\left(9^{3}-\right.$ $\left.7^{3}-2^{3}\right) \cdot 10^{3}=378000=135 \cdot 2800$. For $n \geq 4$, we note that 400 divides $10^{4}$, and $90^{n}+0^{n} \equiv 70^{n}+20^{n}$ $(\bmod 7)$. Therefore $2800 \mid Q(n)$ for all $n$. $\fbox{782}$.

HMMT Nov Guts

HMMT-Nov Guts

AMC

AMC8

8

N/A

What is the value of \[\frac{1}{3}\cdot\frac{2}{4}\cdot\frac{3}{5}\cdots\frac{18}{20}\cdot\frac{19}{21}\cdot\frac{20}{22}?\]

\frac{1}{231}

Note that common factors (from $3$ to $20,$ inclusive) of the numerator and the denominator cancel. Therefore, the original expression becomes \[\frac{1}{\cancel{3}}\cdot\frac{2}{\cancel{4}}\cdot\frac{\cancel{3}}{\cancel{5}}\cdots\frac{\cancel{18}}{\cancel{20}}\cdot\frac{\cancel{19}}{21}\cdot\frac{\cancel{20}}{22}=\frac{1\cdot2}{21\cdot22}=\frac{1}{21\cdot11}=\fbox{\frac{1}{231}}.\]

AMC8 First Half

AMC8

AIME

AIME

I

N/A

Let $ABCD$ be a parallelogram with $\angle BAD < 90^\circ.$ A circle tangent to sides $\overline{DA},$ $\overline{AB},$ and $\overline{BC}$ intersects diagonal $\overline{AC}$ at points $P$ and $Q$ with $AP < AQ,$ as shown. Suppose that $AP=3,$ $PQ=9,$ and $QC=16.$ Then the area of $ABCD$ can be expressed in the form $m\sqrt{n},$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n.$ [asy] defaultpen(linewidth(0.6)+fontsize(11)); size(8cm); pair A,B,C,D,P,Q; A=(0,0); label("$A$", A, SW); B=(6,15); label("$B$", B, NW); C=(30,15); label("$C$", C, NE); D=(24,0); label("$D$", D, SE); P=(5.2,2.6); label("$P$", (5.8,2.6), N); Q=(18.3,9.1); label("$Q$", (18.1,9.7), W); draw(A--B--C--D--cycle); draw(C--A); draw(Circle((10.95,7.45), 7.45)); dot(A^^B^^C^^D^^P^^Q); [/asy]

150

Let's redraw the diagram, but extend some helpful lines. [asy] size(10cm); pair A,B,C,D,EE,F,P,Q,O; A=(0,0); EE = (24,15); F = (30,0); O = (10.5,7.5); label("$A$", A, SW); B=(6,15); label("$B$", B, NW); C=(30,15); label("$C$", C, NE); D=(24,0); label("$D$", D, SE); P=(5.2,2.6); label("$P$", (5.8,2.6), N); Q=(18.3,9.1); label("$Q$", (18.1,9.7), W); draw(A--B--C--D--cycle); draw(C--A); draw(Circle((10.95,7.45), 7.45)); dot(A^^B^^C^^D^^P^^Q); dot(O); label("$O$",O,W); draw((10.5,15)--(10.5,0)); draw(D--(24,15),dashed); draw(C--(30,0),dashed); draw(D--(30,0)); dot(EE); dot(F); label("$3$", midpoint(A--P), S); label("$9$", midpoint(P--Q), S); label("$16$", midpoint(Q--C), S); label("$x$", (5.5,13.75), W); label("$20$", (20.25,15), N); label("$6$", (5.25,0), S); label("$6$", (1.5,3.75), W); label("$x$", (8.25,15),N); label("$14+x$", (17.25,0), S); label("$6-x$", (27,15), N); label("$6+x$", (27,7.5), S); label("$6\sqrt{3}$", (30,7.5), E); label("$T_1$", (10.5,15), N); label("$T_2$", (10.5,0), S); label("$T_3$", (4.5,11.25), W); label("$E$", EE, N); label("$F$", F, S); [/asy] We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let $T_1, T_2, T_3$ be our tangents from the circle to the parallelogram. By the secant power of a point, the power of $A = 3 \cdot (3+9) = 36$. Then $AT_2 = AT_3 = \sqrt{36} = 6$. Similarly, the power of $C = 16 \cdot (16+9) = 400$ and $CT_1 = \sqrt{400} = 20$. We let $BT_3 = BT_1 = x$ and label the diagram accordingly. Notice that because $BC = AD, 20+x = 6+DT_2 \implies DT_2 = 14+x$. Let $O$ be the center of the circle. Since $OT_1$ and $OT_2$ intersect $BC$ and $AD$, respectively, at right angles, we have $T_2T_1CD$ is a right-angled trapezoid and more importantly, the diameter of the circle is the height of the triangle. Therefore, we can drop an altitude from $D$ to $BC$ and $C$ to $AD$, and both are equal to $2r$. Since $T_1E = T_2D$, $20 - CE = 14+x \implies CE = 6-x$. Since $CE = DF, DF = 6-x$ and $AF = 6+14+x+6-x = 26$. We can now use Pythagorean theorem on $\triangle ACF$; we have $26^2 + (2r)^2 = (3+9+16)^2 \implies 4r^2 = 784-676 \implies 4r^2 = 108 \implies 2r = 6\sqrt{3}$ and $r^2 = 27$. We know that $CD = 6+x$ because $ABCD$ is a parallelogram. Using Pythagorean theorem on $\triangle CDF$, $(6+x)^2 = (6-x)^2 + 108 \implies (6+x)^2-(6-x)^2 = 108 \implies 12 \cdot 2x = 108 \implies 2x = 9 \implies x = \frac{9}{2}$. Therefore, base $BC = 20 + \frac{9}{2} = \frac{49}{2}$. Thus the area of the parallelogram is the base times the height, which is $\frac{49}{2} \cdot 6\sqrt{3} = 147\sqrt{3}$ and the answer is $\fbox{150}$ ~KingRavi

Hard AIME Problems

AIME

HMMT

HMMT-Nov

gen

Nov

Let $A B C D$ be a convex trapezoid such that $\angle A B C=\angle B C D=90^{\circ}, A B=3, B C=6$, and $C D=12$. Among all points $X$ inside the trapezoid satisfying $\angle X B C=\angle X D A$, compute the minimum possible value of $C X$.

\sqrt{113}-\sqrt{65}.

Solution: Let $P=A D \cap B C$. Then, we see that the locus if $X$ is the $\operatorname{arc} \widehat{B D}$ of $\odot(P B D)$. Let $O$ be its center, and $R$ be the radius of $\odot(P B D)$, then the answer is $C O-R$. Let $T$ be the second intersection of $\odot(P B D)$ and $C D$. We can compute $B P=2$, so by Power of Point, $C T \cdot C D=C P \cdot C B=48$, so $C T=4$. Since the projections of $O$ onto $C D$ and $C B$ are midpoints of $B P$ and $D T$, we get by Pythagorean theorem that $C O=\sqrt{8^{2}+7^{2}}=\sqrt{113}$. By Pythagorean theorem, we also get that $R=\sqrt{8^{2}+1^{2}}=\sqrt{65}$, hence the answer. $\fbox{\sqrt{113}-\sqrt{65}.}$.

HMMT Nov Hard

HMMT-Nov General

HMMT

HMMT-Nov

thm

Nov

Alice is bored in class, so she thinks of a positive integer. Every second after that, she subtracts from her current number its smallest prime divisor, possibly itself. After 2022 seconds, she realizes that her number is prime. Find the sum of all possible values of her initial number.

8093

Solution: Let $a_{k}$ denote Alice's number after $k$ seconds, and let $p_{k}$ be the smallest prime divisor of $a_{k}$. We are given that $a_{2022}$ is prime, and want to find $a_{0}$. If $a_{0}$ is even, then $a_{n+1}=a_{n}-2$, since every $a_{n}$ is even. Then we need $a_{2022}=2$, so $a_{0}=4046$. If $a_{0}$ is odd, then $a_{1}=a_{0}-p_{0}$ is even, so by similar logic to the even case, $a_{1}=4044$. Then since $p_{0} \mid a_{0}-p_{0}$ and $4044=4 \cdot 3 \cdot 337$, we must have $p_{0}=3$ or 337 . But if $p_{0}=337, a_{0}=12 \cdot 337+337=13 \cdot 337$, so 337 is not the smallest prime divisor of $a_{0}$. Thus, we need $p_{0}=3$, so $a_{0}=4047$, which works. Thus, the final answer is $4046+4047=8093$. $\fbox{8093}$.

HMMT Nov Easy

HMMT-Nov Theme

HMMT

HMMT-Nov

team

Nov

Let $A B C D$ be a convex trapezoid such that $\angle B A D=\angle A D C=90^{\circ}, A B=20, A D=21$, and $C D=28$. Point $P \neq A$ is chosen on segment $A C$ such that $\angle B P D=90^{\circ}$. Compute $A P$.

9

Solution 1: Construct the rectangle $A B X D$. Note that \[ \angle B A D=\angle B P D=\angle B X D=90^{\circ} \] so $A B X P D$ is cyclic with diameter $B D$. By Power of a Point, we have $C X \cdot C D=C P \cdot C A$. Note that $C X=C D-X D=C D-A B=8$ and $C A=\sqrt{A D^{2}+D C^{2}}=35$. Therefore, \[ C P=\frac{C X \cdot C D}{C A}=\frac{8 \cdot 28}{35}=\frac{32}{5} \] and so \[ A P=A C-C P=35-\frac{32}{5}=\frac{143}{5} \] Solution 2: Since $A B P D$ is cyclic and $\angle B D P=\angle B A P=\angle A C D$, it follows that $\triangle B P D \sim \triangle A B C$. Thus, if $B D=5 x, B P=4 x$, and $D P=3 x$, then by Ptolemy's theorem, we have \[ \begin{aligned} A P \cdot(5 x) & =A D \cdot(3 x)+A B \cdot(4 x) \\ A P & =\frac{21 \cdot 3+20 \cdot 4}{5}=\frac{143}{5} \end{aligned} \] $\fbox{9}$.

HMMT Nov Team

HMMT-Nov Team

AMC

AMC10

10B

N/A

A $2 \times 3$ rectangle and a $3 \times 4$ rectangle are contained within a square without overlapping at any point, and the sides of the square are parallel to the sides of the two given rectangles. What is the smallest possible area of the square?

25

By placing the $2 \times 3$ rectangle adjacent to the $3 \times 4$ rectangle with the 3 side of the $2 \times 3$ rectangle next to the 4 side of the $3 \times 4$ rectangle, we get a figure that can be completely enclosed in a square with a side length of 5. The area of this square is $5^2 = 25$. Since placing the two rectangles inside a $4 \times 4$ square must result in overlap, the smallest possible area of the square is $25$. So the answer is $\fbox{25}$.

AMC10 First Half

AMC10 B

AIME

AIME

I

N/A

For each integer $n \ge 2$, let $A(n)$ be the area of the region in the coordinate plane defined by the inequalities $1\le x \le n$ and $0\le y \le x \left\lfloor \sqrt x \right\rfloor$, where $\left\lfloor \sqrt x \right\rfloor$ is the greatest integer not exceeding $\sqrt x$. Find the number of values of $n$ with $2\le n \le 1000$ for which $A(n)$ is an integer.

483

Let $n\ge 2$ and define $a(n) = \left\lfloor \sqrt n \right\rfloor$. For $2\le n \le 1000$, we have $1\le a(n)\le 31$. For $a^2 \le x < (a+1)^2$ we have $y=ax$. Thus $A(n+1)-A(n)=a(n+\tfrac 12) = \Delta_n$ (say), and $\Delta_n$ is an integer if $a$ is even; otherwise $\Delta_n$ is an integer plus $\tfrac 12$. If $a=1$, $n\in \{1,2,3\}$ and $\Delta_n$ is of the form $k(n)+\tfrac 12$ so $A(n)$ is an integer when $n$ is even. If $a=2$, $n\in\{4,\ldots , 8\}$ and $\Delta_n$ is an integer for all $n$. Since $A(3)$ is not an integer, so $A(n)$ is not an integer for any $n$. If $a=3$, $n\in\{9,\ldots , 15\}$ and $\Delta_n$ is of the form $k(n)+\tfrac 12$. Since $A(8)$ is of the form $k+\tfrac 12$ so $A(n)$ is an integer only when $n$ is odd. If $a=4$, $n\in\{16,\ldots , 24\}$ and $\Delta_n$ is an integer for all $n$. Since $A(15)$ is an integer so $A(n)$ is an integer for all $n$. Now we are back to where we started; i.e., the case $a=5$ will be the same as $a=1$ and so on. Thus, \begin{align} a(n)\equiv 1\pmod 4 \qquad &\Longrightarrow \qquad A(n) \in \mathbb{Z} \textrm{ for even } n, \\ a(n)\equiv 2\pmod 4 \qquad &\Longrightarrow \qquad A(n) \not\in \mathbb{Z} \textrm{ for any } n, \\ a(n)\equiv 3\pmod 4 \qquad &\Longrightarrow \qquad A(n) \in \mathbb{Z} \textrm{ for odd } n, \\ a(n)\equiv 0\pmod 4 \qquad &\Longrightarrow \qquad A(n) \in \mathbb{Z} \textrm{ for all } n. \end{align} For each $a$ there are $2a+1$ corresponding values of $n$: i.e., $n\in \{a^2, \ldots , (a+1)^2-1\}$. Thus, the number of values of $n$ corresponding to $(4)$ (i.e., $a(n)\equiv 0\pmod 4$) is given by \[\sum_{\substack{a=4k \\ a\le 31}}(2a+1) = \sum_{k=1}^7 (8k+1)=231.\] The cases $(1)$ and $(3)$ combine to account for half the values of $n$ corresponding to odd values of $a(n)$; i.e., \[\frac 12 \cdot \sum_{\substack{a=2k+1 \\ a\le 31}} (2a+1) = \sum_{k=0}^{15} (2k+\tfrac 32) = 264\]However, this also includes the odd integers in $\{1001, \ldots , 1023\}$. Subtracting $12$ to account for these, we get the number of values of $n$ corresponding to cases $(1)$ and $(3)$ to be $264-12=252$. Adding the contributions from all cases we get our answer to be $231+252= \fbox{483}$.

Very Hard AIME Problems

AIME

AMC

AMC10

10A

N/A

The polynomial $x^3-ax^2+bx-2010$ has three positive integer roots. What is the smallest possible value of $a$?

78

We can expand $(x+a)(x+b)(x+c)$ as $(x^2+ax+bx+ab)(x+c)$ $(x^2+ax+bx+ab)(x+c)=x^3+abx+acx+bcx+abx^2+acx^2+bcx^2+abc=x^3+x^2(a+b+c)+x(ab+ac+bc)+abc$ We do not care about $+bx$ in this case, because we are only looking for $a$. We know that the constant term is $-2010=-(2\cdot 3\cdot 5\cdot 67)$ We are trying to minimize a, such that we have $-ax^2$ Since we have three positive solutions, we have $(x-a)(x-b)(x-c)$ as our factors. We have to combine two of the factors of $2\cdot 3\cdot 5\cdot 67$, and then sum up the $3$ resulting factors. Since we are minimizing, we choose $2$ and $3$ to combine together. We get $(x-6)(x-5)(x-67)$ which gives us a coefficient of $x^2$ of $-6-5-67=-78$ Therefore $-a=-78$ or $a=\fbox{78}$

AMC10 Final Problems

AMC10 A

HMMT

HMMT-Feb

guts

Feb

Let $N$ be a three-digit integer such that the difference between any two positive integer factors of $N$ is divisible by 3 . Let $d(N)$ denote the number of positive integers which divide $N$. Find the maximum possible value of $N \cdot d(N)$.

5586

We first note that all the prime factors of $n$ must be 1 modulo 3 (and thus 1 modulo 6). The smallest primes with this property are $7,13,19, \ldots$ Since $7^{4}=2401>1000$, the number can have at most 3 prime factors (including repeats). Since $7 \cdot 13 \cdot 19=1729>1000$, the most factors $N$ can have is 6 . Consider the number $7^{2} \cdot 19=931$, which has 6 factors. For this choice of $N, N \cdot d(N)=5586$. For another $N$ to do better, it must have at least 6 factors, for otherwise, $N \cdot d(N)<1000 \cdot 5=5000$. It is easy to verify that $7^{2} \cdot 19$ is the greatest number with 6 prime factors satisfying our conditions, so the answer must be 5586 . $\fbox{5586}$.

HMMT Feb Guts

HMMT-Feb Guts

AMC

AMC8

8

N/A

A block wall 100 feet long and 7 feet high will be constructed using blocks that are 1 foot high and either 2 feet long or 1 foot long (no blocks may be cut). The vertical joins in the blocks must be staggered as shown, and the wall must be even on the ends. What is the smallest number of blocks needed to build this wall? [asy] draw((0,0)--(6,0)--(6,1)--(5,1)--(5,2)--(0,2)--cycle); draw((0,1)--(5,1)); draw((1,1)--(1,2)); draw((3,1)--(3,2)); draw((2,0)--(2,1)); draw((4,0)--(4,1)); [/asy]

353

Since the bricks are $1$ foot high, there will be $7$ rows. To minimize the number of blocks used, rows $1, 3, 5,$ and $7$ will look like the bottom row of the picture, which takes $\frac{100}{2} = 50$ bricks to construct. Rows $2, 4,$ and $6$ will look like the upper row pictured, which has $49$ 2-foot bricks in the middle, and $2$ 1-foot bricks on each end for a total of $51$ bricks. Four rows of $50$ bricks and three rows of $51$ bricks totals $4\cdot 50 + 3\cdot 51 = 200 + 153 = 353$ bricks, giving the answer $\fbox{353}.$

AMC8 First Half

AMC8

AMC

AMC12

12A

N/A

A fancy bed and breakfast inn has $5$ rooms, each with a distinctive color-coded decor. One day $5$ friends arrive to spend the night. There are no other guests that night. The friends can room in any combination they wish, but with no more than $2$ friends per room. In how many ways can the innkeeper assign the guests to the rooms?

2220

We can discern three cases. Case 1: Each room houses one guest. In this case, we have $5$ guests to choose for the first room, $4$ for the second, ..., for a total of $5!=120$ assignments. Case 2: Three rooms house one guest; one houses two. We have $\binom{5}{3}$ ways to choose the three rooms with $1$ guest, and $\binom{2}{1}$ to choose the remaining one with $2$. There are $5\cdot4\cdot3$ ways to place guests in the first three rooms, with the last two residing in the two-person room, for a total of $\binom{5}{3}\binom{2}{1}\cdot5\cdot4\cdot3=1200$ ways. Case 3: Two rooms house two guests; one houses one. We have $\binom{5}{2}$ to choose the two rooms with two people, and $\binom{3}{1}$ to choose one remaining room for one person. Then there are $5$ choices for the lonely person, and $\binom{4}{2}$ for the two in the first two-person room. The last two will stay in the other two-room, so there are $\binom{5}{2}\binom{3}{1}\cdot5\cdot\binom{4}{2}=900$ ways. In total, there are $120+1200+900=2220$ assignments, or $\fbox{2220}$. (Solution by AwesomeToad16)

AMC12 Second Half

AMC12 A

AMC

AMC12

12B

N/A

In $\triangle ABC$, $\cos(2A-B)+\sin(A+B)=2$ and $AB=4$. What is $BC$?

2

We note that $-1$ $\le$ $\sin x$ $\le$ $1$ and $-1$ $\le$ $\cos x$ $\le$ $1$. Therefore, there is no other way to satisfy this equation other than making both $\cos(2A-B)=1$ and $\sin(A+B)=1$, since any other way would cause one of these values to become greater than 1, which contradicts our previous statement. From this we can easily conclude that $2A-B=0^{\circ}$ and $A+B=90^{\circ}$ and solving this system gives us $A=30^{\circ}$ and $B=60^{\circ}$. It is clear that $\triangle ABC$ is a $30^{\circ},60^{\circ},90^{\circ}$ triangle with $BC=2$　$\Longrightarrow$　$(C)$. $\fbox{2}$.

AMC12 Second Half

AMC12 B

AIME

AIME

II

N/A

A frog is positioned at the origin of the coordinate plane. From the point $(x, y)$, the frog can jump to any of the points $(x + 1, y)$, $(x + 2, y)$, $(x, y + 1)$, or $(x, y + 2)$. Find the number of distinct sequences of jumps in which the frog begins at $(0, 0)$ and ends at $(4, 4)$.

556

We solve this problem by working backwards. Notice, the only points the frog can be on to jump to $(4,4)$ in one move are $(2,4),(3,4),(4,2),$ and $(4,3)$. This applies to any other point, thus we can work our way from $(0,0)$ to $(4,4)$, recording down the number of ways to get to each point recursively. $(0,0): 1$ $(1,0)=(0,1)=1$ $(2,0)=(0, 2)=2$ $(3,0)=(0, 3)=3$ $(4,0)=(0, 4)=5$ $(1,1)=2$, $(1,2)=(2,1)=5$, $(1,3)=(3,1)=10$, $(1,4)=(4,1)= 20$ $(2,2)=14, (2,3)=(3,2)=32, (2,4)=(4,2)=71$ $(3,3)=84, (3,4)=(4,3)=207$ $(4,4)=2\cdot \left( (4,2)+(4,3)\right) = 2\cdot \left( 207+71\right)=2\cdot 278=\fbox{556}$ A diagram of the numbers: [asy] import graph; add(shift(0,0)*grid(4,4)); label((0,0), "1", SW); label((1,0), "1", SW); label((2,0), "2", SW); label((3,0), "3", SW); label((4,0), "5", SW); label((0,1), "1", SW); label((1,1), "2", SW); label((2,1), "5", SW); label((3,1), "10", SW); label((4,1), "20", SW); label((0,2), "2", SW); label((1,2), "5", SW); label((2,2), "14", SW); label((3,2), "32", SW); label((4,2), "71", SW); label((0,3), "3", SW); label((1,3), "10", SW); label((2,3), "32", SW); label((3,3), "84", SW); label((4,3), "207", SW); label((0,4), "5", SW); label((1,4), "20", SW); label((2,4), "71", SW); label((3,4), "207", SW); label((4,4), "556", SW); [/asy] ~First

Intermediate AIME Problems

AIME

AMC

AMC10

10A

N/A

A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with the terms 247, 475, and 756 and end with the term 824. Let $S$ be the sum of all the terms in the sequence. What is the largest prime factor that always divides $S$?

37

A given digit appears as the hundreds digit, the tens digit, and the units digit of a term the same number of times. Let $k$ be the sum of the units digits in all the terms. Then $S=111k=3 \cdot 37k$, so $S$ must be divisible by $37\ \mathrm{(D)}$. To see that it need not be divisible by any larger prime, the sequence $123, 231, 312$ gives $S=666=2 \cdot 3^2 \cdot 37\Rightarrow \mathrm{\fbox{37}}$.

AMC10 Final Problems

AMC10 A

HMMT

HMMT-Nov

team

Nov

$A B C$ is a triangle with $A B=15, B C=14$, and $C A=13$. The altitude from $A$ to $B C$ is extended to meet the circumcircle of $A B C$ at $D$. Find $A D$.

\frac{63}{4}

Let the altitude from $A$ to $B C$ meet $B C$ at $E$. The altitude $A E$ has length 12; one way to see this is that it splits the triangle $A B C$ into a $9-12-15$ right triangle and a $5-12-13$ right triangle; from this, we also know that $B E=9$ and $C E=5$. Now, by Power of a Point, $A E \cdot D E=B E \cdot C E$, so $D E=(B E \cdot C E) / A E=(9 \cdot 5) /(12)=15 / 4$. It then follows that $A D=A E+D E=63 / 4$. $\fbox{\frac{63}{4}}$.

HMMT Nov Team

HMMT-Nov Team

AIME

AIME

II

N/A

Let $A = \{1, 2, 3, 4, 5, 6, 7\}$, and let $N$ be the number of functions $f$ from set $A$ to set $A$ such that $f(f(x))$ is a constant function. Find the remainder when $N$ is divided by $1000$.

399

Any such function can be constructed by distributing the elements of $A$ on three tiers. The bottom tier contains the constant value, $c=f(f(x))$ for any $x$. (Obviously $f(c)=c$.) The middle tier contains $k$ elements $x\ne c$ such that $f(x)=c$, where $1\le k\le 6$. The top tier contains $6-k$ elements such that $f(x)$ equals an element on the middle tier. There are $7$ choices for $c$. Then for a given $k$, there are $\tbinom6k$ ways to choose the elements on the middle tier, and then $k^{6-k}$ ways to draw arrows down from elements on the top tier to elements on the middle tier. Thus $N=7\cdot\sum_{k=1}^6\tbinom6k\cdot k^{6-k}=7399$, giving the answer $\fbox{399}$.

Hard AIME Problems

AIME

AMC

AMC10

10B

N/A

An annulus is the region between two concentric circles. The concentric circles in the figure have radii $b$ and $c$, with $b>c$. Let $OX$ be a radius of the larger circle, let $XZ$ be tangent to the smaller circle at $Z$, and let $OY$ be the radius of the larger circle that contains $Z$. Let $a=XZ$, $d=YZ$, and $e=XY$. What is the area of the annulus? [asy] unitsize(1.5cm); defaultpen(0.8); real r1=1.5, r2=2.5; pair O=(0,0); path inner=Circle(O,r1), outer=Circle(O,r2); pair Y=(0,r2), Z=(0,r1), X=intersectionpoint( Z--(Z+(10,0)), outer ); filldraw(outer,lightgray,black); filldraw(inner,white,black); draw(X--O--Y); draw(Y--X--Z); label("$O$",O,SW); label("$X$",X,E); label("$Y$",Y,N); label("$Z$",Z,SW); label("$a$",X--Z,N); label("$b$",0.25*X,SE); label("$c$",O--Z,E); label("$d$",Y--Z,W); label("$e$",Y*0.65 + X*0.35,SW); defaultpen(0.5); dot(O); dot(X); dot(Z); dot(Y); [/asy]

\pi a^2

The area of the large circle is $\pi b^2$, the area of the small one is $\pi c^2$, hence the shaded area is $\pi(b^2-c^2)$. From the Pythagorean Theorem for the right triangle $OXZ$ we have $a^2 + c^2 = b^2$, hence $b^2-c^2=a^2$ and thus the shaded area is $\fbox{\pi a^2}$.

AMC10 Second Half

AMC10 B

HMMT

HMMT-Nov

gen

Nov

Five equally skilled tennis players named Allen, Bob, Catheryn, David, and Evan play in a round robin tournament, such that each pair of people play exactly once, and there are no ties. In each of the ten games, the two players both have a $50 \%$ chance of winning, and the results of the games are independent. Compute the probability that there exist four distinct players $P_{1}, P_{2}, P_{3}, P_{4}$ such that $P_{i}$ beats $P_{i+1}$ for $i=1,2,3,4$. (We denote $P_{5}=P_{1}$ ).

\frac{49}{64}

We make the following claim: if there is a 5-cycle (a directed cycle involving 5 players) in the tournament, then there is a 4-cycle. Proof: Assume that $A$ beats $B, B$ beats $C, C$ beats $D, D$ beats $E$ and $E$ beats $A$. If $A$ beats $C$ then $A, C, D, E$ forms a 4 -cycle, and similar if $B$ beats $D, C$ beats $E$, and so on. However, if all five reversed matches occur, then $A, D, B, C$ is a 4 -cycle. Therefore, if there are no 4-cycles, then there can be only 3-cycles or no cycles at all. Case 1: There is a 3-cycle. Assume that $A$ beats $B, B$ beats $C$, and $C$ beats $A$. (There are $\left(\begin{array}{l}5 \\ 3\end{array}\right)=10$ ways to choose the cycle and 2 ways to orient the cycle.) Then $D$ either beats all three or is beaten by all three, because otherwise there exists two people $X$ and $Y$ in these three people such that $X$ beats $Y$, and $D$ beats $Y$ but is beaten by $X$, and then $X, D, Y, Z$ will form a 4-cycle ( $Z$ is the remaining\\ person of the three). The same goes for $E$. If $D$ and $E$ both beat all three or are beaten by all three, then there is no restriction on the match between $D$ and $E$. However, if $D$ beats all three and $E$ loses to all three, then $E$ cannot beat $D$ because otherwise $E, D, A, B$ forms a 4 -cycle. This means that $A, B, C$ is the only 3 -cycle in the tournament, and once the cycle is chosen there are $2 \cdot 2+2 \cdot 1=6$ ways to choose the results of remaining matches, for $10 \cdot 2 \cdot 6=120$ ways in total. Case 2: There are no cycles. This means that the tournament is a complete ordering (the person with a higher rank always beats the person with a lower rank). There are $5 !=120$ ways in this case as well. Therefore, the probability of not having a 4 -cycle is $\frac{120+120}{2^{10}}=\frac{15}{64}$, and thus the answer is $1-\frac{15}{64}=\frac{49}{64}$. $\fbox{\frac{49}{64}}$.

HMMT Nov Hard

HMMT-Nov General

HMMT

HMMT-Nov

gen

Nov

In $\triangle A B C, A B=2019, B C=2020$, and $C A=2021$. Yannick draws three regular $n$-gons in the plane of $\triangle A B C$ so that each $n$-gon shares a side with a distinct side of $\triangle A B C$ and no two of the $n$-gons overlap. What is the maximum possible value of $n$ ?

11

If any $n$-gon is drawn on the same side of one side of $\triangle A B C$ as $\triangle A B C$ itself, it will necessarily overlap with another triangle whenever $n>3$. Thus either $n=3$ or the triangles are all outside $A B C$. The interior angle of a regular $n$-gon is $180^{\circ} \cdot \frac{n-2}{n}$, so we require \[ 360^{\circ} \cdot \frac{n-2}{n}+\max (\angle A, \angle B, \angle C)<360^{\circ} \] As $\triangle A B C$ is almost equilateral (in fact the largest angle is less than $60.1^{\circ}$ ), each angle is approximately $60^{\circ}$, so we require \[ 360 \cdot \frac{n-2}{n}<300 \Longrightarrow n<12 \] Hence the answer is $n=11$. $\fbox{11}$.

HMMT Nov Easy

HMMT-Nov General

AMC

AMC10

10A

N/A

Bernardo randomly picks 3 distinct numbers from the set $\{1,2,3,4,5,6,7,8,9\}$ and arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set $\{1,2,3,4,5,6,7,8\}$ and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo's number is larger than Silvia's number?

\frac{37}{56}

We can solve this by breaking the problem down into $2$ cases and adding up the probabilities. Case $1$: Bernardo picks $9$. If Bernardo picks a $9$ then it is guaranteed that his number will be larger than Silvia's. The probability that he will pick a $9$ is $\frac{1 \cdot \binom{8}{2}}{\binom{9}{3}} = \frac{\frac{8\cdot7}{2}}{\frac{9\cdot8\cdot7}{3\cdot2\cdot1}}=\frac{1}{3}$. Case $2$: Bernardo does not pick $9$. Since the chance of Bernardo picking $9$ is $\frac{1}{3}$, the probability of not picking $9$ is $\frac{2}{3}$. If Bernardo does not pick $9$, then he can pick any number from $1$ to $8$. Since Bernardo is picking from the same set of numbers as Silvia, the probability that Bernardo's number is larger is equal to the probability that Silvia's number is larger. Ignoring the $9$ for now, the probability that they will pick the same number is the number of ways to pick Bernardo's 3 numbers divided by the number of ways to pick any 3 numbers. We get this probability to be $\frac{3!}{8\cdot{7}\cdot{6}} = \frac{1}{56}$ The probability of Bernardo's number being greater is \[\frac{1-\frac{1}{56}}{2} = \frac{55}{112}\] Factoring the fact that Bernardo could've picked a $9$ but didn't: \[\frac{2}{3}\cdot{\frac{55}{112}} = \frac{55}{168}\] Adding up the two cases we get $\frac{1}{3}+\frac{55}{168} = \fbox{\frac{37}{56}}$

AMC10 Second Half

AMC10 A

HMMT

HMMT-Feb

guts

Feb

Five people take a true-or-false test with five questions. Each person randomly guesses on every question. Given that, for each question, a majority of test-takers answered it correctly, let $p$ be the probability that every person answers exactly three questions correctly. Suppose that $p=\frac{a}{2^{b}}$ where $a$ is an odd positive integer and $b$ is a nonnegative integer. Compute $100 a+b$.

25517

Solution: There are a total of $16^{5}$ ways for the people to collectively ace the test. Consider groups of people who share the same problems that they got incorrect. We either have a group of 2 and a group of 3 , or a group 5 . In the first case, we can pick the group of two in $\left(\begin{array}{l}5 \\ 2\end{array}\right)$ ways, the problems they got wrong in $\left(\begin{array}{l}5 \\ 2\end{array}\right)$ ways. Then there are 3 ! ways for the problems of group 3. There are 600 cases here. In the second case, we can $5 ! \cdot 4 ! / 2=120 \cdot 12$ ways to organize the five cycle ( $4 ! / 2$ to pick a cycle and 5 ! ways to assign a problem to each edge in the cycle). Thus, the solution is $\frac{255}{2^{17}}$ and the answer is 25517 . $\fbox{25517}$.

HMMT Feb Guts

HMMT-Feb Guts

HMMT

HMMT-Nov

guts

Nov

Let $A B C$ be a triangle with $A B=13, B C=14$, and $C A=15$. We construct isosceles right triangle $A C D$ with $\angle A D C=90^{\circ}$, where $D, B$ are on the same side of line $A C$, and let lines $A D$ and $C B$ meet at $F$. Similarly, we construct isosceles right triangle $B C E$ with $\angle B E C=90^{\circ}$, where $E, A$ are on the same side of line $B C$, and let lines $B E$ and $C A$ meet at $G$. Find $\cos \angle A G F$.

-\frac{5}{13}

We see that $\angle G A F=\angle G B F=45^{\circ}$, hence quadrilateral $G F B A$ is cyclic. Consequently $\angle A G F+\angle F B A=180^{\circ}$. So $\cos \angle A G F=-\cos \angle F B A$. One can check directly that $\cos \angle C B A=\frac{5}{13}$ (say, by the Law of Cosines). $\fbox{-\frac{5}{13}}$.

HMMT Nov Guts

HMMT-Nov Guts

AIME

AIME

I

N/A

Let $S$ be the set of all rational numbers that can be expressed as a repeating decimal in the form $0.\overline{abcd},$ where at least one of the digits $a,$ $b,$ $c,$ or $d$ is nonzero. Let $N$ be the number of distinct numerators obtained when numbers in $S$ are written as fractions in lowest terms. For example, both $4$ and $410$ are counted among the distinct numerators for numbers in $S$ because $0.\overline{3636} = \frac{4}{11}$ and $0.\overline{1230} = \frac{410}{3333}.$ Find the remainder when $N$ is divided by $1000.$

392

$0.\overline{abcd}=\frac{abcd}{9999} = \frac{x}{y}$, $9999=9\times 11\times 101$. Then we need to find the number of positive integers $x$ that (with one of more $y$ such that $y|9999$) can meet the requirement $1 \leq {x}\cdot\frac{9999}{y} \leq 9999$. Make cases by factors of $x$. (A venn diagram of cases would be nice here.) Case $A$: $3 \nmid x$ and $11 \nmid x$ and $101 \nmid x$, aka $\gcd (9999, x)=1$. Euler's totient function counts these: \[\varphi \left(3^2 \cdot 11 \cdot 101 \right) = ((3-1)\cdot 3)(11-1)(101-1)= \bf{6000}\] values (but it's enough to note that it's a multiple of 1000 and thus does not contribute to the final answer) Note: You don't need to know this formula. The remaining cases essentially re-derive the same computation for other factors of $9999$. This case isn't actually different. The remaining cases have $3$ (or $9$), $11$, and/or $101$ as factors of $abcd$, which cancel out part of $9999$. Note: Take care about when to use $3$ vs $9$. Case $B$: $3|x$, but $11 \nmid x$ and $101 \nmid x$. Then $abcd=9x$ to leave 3 uncancelled, and $x=3p$, so $x \leq \frac{9999}{9} = 1111$, giving: $x \in 3 \cdot \{1, \dots \left\lfloor \frac{1111}{3}\right\rfloor\}$, $x \notin (3\cdot 11) \cdot \{1 \dots \left\lfloor \frac{1111}{3\cdot 11}\right\rfloor\}$, $x \notin (3 \cdot 101) \cdot \{1 \dots \left\lfloor \frac{1111}{3 \cdot 101}\right\rfloor\}$, for a subtotal of $\left\lfloor \frac{1111}{3}\right\rfloor - (\left\lfloor\frac{1111}{3 \cdot 11}\right\rfloor + \left\lfloor\frac{1111}{3 \cdot 101}\right\rfloor ) = 370 - (33+3) = \bf{334}$ values. Case $C$: $11|x$, but $3 \nmid x$ and $101 \nmid x$. Much like previous case, $abcd$ is $11x$, so $x \leq \frac{9999}{11} = 909$, giving $\left\lfloor \frac{909}{11}\right\rfloor - \left(\left\lfloor\frac{909}{11 \cdot 3}\right\rfloor + \left\lfloor\frac{909}{11 \cdot 101}\right\rfloor \right) = 82 - (27 + 0) = \bf{55}$ values. Case $D$: $3|x$ and $11|x$ (so $33|x$), but $101 \nmid x$. Here, $abcd$ is $99x$, so $x \leq \frac{9999}{99} = 101$, giving $\left\lfloor \frac{101}{33}\right\rfloor - \left\lfloor \frac{101}{33 \cdot 101}\right\rfloor = 3-0 = \bf{3}$ values. Case $E$: $101|x$. Here, $abcd$ is $101x$, so $x \leq \frac{9999}{101} = 99$, giving $\left\lfloor \frac{99}{101}\right\rfloor = \bf{0}$ values, so we don't need to account for multiples of $3$ and $11$. To sum up, the answer is \[6000+334+55+3+0\equiv\fbox{392} \pmod{1000}.\]

Very Hard AIME Problems

AIME

HMMT

HMMT-Nov

guts

Nov

A circle of radius 6 is drawn centered at the origin. How many squares of side length 1 and integer coordinate vertices intersect the interior of this circle?

132

By symmetry, the answer is four times the number of squares in the first quadrant. Let's identify each square by its coordinates at the bottom-left corner, $(x, y)$. When $x=0$, we can have $y=0 \ldots 5$, so there are 6 squares. (Letting $y=6$ is not allowed because that square intersects only the boundary of the circle.) When $x=1$, how many squares are there? The equation of the circle is $y=\sqrt{36-x^{2}}=\sqrt{36-1^{2}}$ is between 5 and 6 , so we can again have $y=0 \ldots 5$. Likewise for $x=2$ and $x=3$. When $x=4$ we have $y=\sqrt{20}$ which is between 4 and 5 , so there are 5 squares, and when $x=5$ we have $y=\sqrt{11}$ which is between 3 and 4 , so there are 4 squares. Finally, when $x=6$, we have $y=0$, and no squares intersect the interior of the circle. This gives $6+6+6+6+5+4=33$. Since this is the number in the first quadrant, we multiply by four to get 132 . $\fbox{132}$.

HMMT Nov Guts

HMMT-Nov Guts

HMMT

HMMT-Feb

geo

Feb

In quadrilateral $A B C D$, we have $A B=5, B C=6, C D=5, D A=4$, and $\angle A B C=90^{\circ}$. Let $A C$ and $B D$ meet at $E$. Compute $\frac{B E}{E D}$.

\sqrt{3}

We find that $A C=\sqrt{61}$, and applying the law of cosines to triangle $A C D$ tells us that $\angle A D C=120$. Then $\frac{B E}{E D}$ is the ratio of the areas of triangles $A B C$ and $A D C$, which is $\frac{(5)(6)}{(4)(5) \frac{\sqrt{3}}{2}}=\sqrt{3}$. $\fbox{\sqrt{3}}$.

HMMT Feb Hard

HMMT-Feb Geometry

HMMT

HMMT-Feb

comb

Feb

The numbers $1,2, \ldots, 10$ are randomly arranged in a circle. Let $p$ be the probability that for every positive integer $k<10$, there exists an integer $k^{\prime}>k$ such that there is at most one number between $k$ and $k^{\prime}$ in the circle. If $p$ can be expressed as $\frac{a}{b}$ for relatively prime positive integers $a$ and $b$, compute $100 a+b$.

1390

Solution: Let $n=10$ and call two numbers close if there is at most one number between them and an circular permutation focused if only $n$ is greater than all numbers close to it. Let $A_{n}$ be the number of focused circular permutations of $\{1,2, \ldots, n\}$. If $n \geq 5$, then there are 2 cases: $n-1$ is either one or two positions from $n$. If $n-1$ is one position from $n$, it is either on its left or right. In this case, one can check a permutation is focused if and only if removing $n$ yields a focused permutation, so there are $2 A_{n-1}$ permutations in this case. If $n-1$ is two positions from $n$, there are $n-2$ choices for $k$, the element that lies between $n$ and $n-1$. One\\ can show that this permutation is focused if and only if removing both $n$ and $k$ and relabeling the numbers yields a focused permutation, so there are $2(n-2) A_{n-2}$ permutations in this case. Thus, we have $A_{n}=2 A_{n-1}+2(n-2) A_{n-2}$. If we let $p_{n}=A_{n} /(n-1)$ ! the probability that a random circular permutation is focused, then this becomes \[ p_{n}=\frac{2 p_{n-1}+2 p_{n-2}}{n-1} \] Since $p_{3}=p_{4}=1$, we may now use this recursion to calculate \[ p_{5}=1, p_{6}=\frac{4}{5}, p_{7}=\frac{3}{5}, p_{8}=\frac{2}{5}, p_{9}=\frac{1}{4}, p_{10}=\frac{13}{90} \] $\fbox{1390}$.

HMMT Feb Hard

HMMT-Feb Combinatorics

AMC

AMC10

10A

N/A

Let $a$ and $b$ be relatively prime positive integers with $a>b>0$ and $\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}$. What is $a-b$?

3

Since $a$ and $b$ are relatively prime, $a^3-b^3$ and $(a-b)^3$ are both integers as well. Then, for the given fraction to simplify to $\frac{73}{3}$, the denominator $(a-b)^3$ must be a multiple of $3.$ Thus, $a-b$ is a multiple of $3$. Looking at the answer choices, the only multiple of $3$ is $\fbox{3}$.

AMC10 Second Half

AMC10 A

AMC

AMC12

12A

N/A

The numbers $\log(a^3b^7)$, $\log(a^5b^{12})$, and $\log(a^8b^{15})$ are the first three terms of an arithmetic sequence, and the $12^\text{th}$ term of the sequence is $\log{b^n}$. What is $n$?

112

Let $A = \log(a)$ and $B = \log(b)$. The first three terms of the arithmetic sequence are $3A + 7B$, $5A + 12B$, and $8A + 15B$, and the $12^\text{th}$ term is $nB$. Thus, $2(5A + 12B) = (3A + 7B) + (8A + 15B) = 2B$. Since the first three terms in the sequence are $13B$, $22B$, and $31B$, the $k$th term is $(9k + 4)B$. Thus the $12^\text{th}$ term is $(9\cdot12 + 4)B = 112B = nB \Rightarrow n = 112\Rightarrow \fbox{112}$.

AMC12 Second Half

AMC12 A

AMC

AMC10

10B

N/A

Let $n$ be a positive integer greater than 4 such that the decimal representation of $n!$ ends in $k$ zeros and the decimal representation of $(2n)!$ ends in $3k$ zeros. Let $s$ denote the sum of the four least possible values of $n$. What is the sum of the digits of $s$?

8

A trailing zero requires a factor of two and a factor of five. Since factors of two occur more often than factors of five, we can focus on the factors of five. We make a chart of how many trailing zeros factorials have: \[\begin{array}{c|c|c|c|c|c|c} \mathrm{Factorial}&0!-4!&5!-9!&10!-14!&15!-19!&20!-24!&25!-29!\\\hline \mathrm{Zeros}&0&1&2&3&4&6 \end{array}\] We first look at the case when $n!$ has $1$ zero and $(2n)!$ has $3$ zeros. If $n=5,6,7$, $(2n)!$ has only $2$ zeros. But for $n=8,9$, $(2n)!$ has $3$ zeros. Thus, $n=8$ and $n=9$ work. Secondly, we look at the case when $n!$ has $2$ zeros and $(2n)!$ has $6$ zeros. If $n=10,11,12$, $(2n)!$ has only $4$ zeros. But for $n=13,14$, $(2n)!$ has $6$ zeros. Thus, the smallest four values of $n$ that work are $n=8,9,13,14$, which sum to $44$. The sum of the digits of $44$ is $\fbox{8}$

AMC10 Final Problems

AMC10 B

HMMT

HMMT-Feb

alg

Feb

Let $f$ be a monic cubic polynomial satisfying $f(x)+f(-x)=0$ for all real numbers $x$. For all real numbers $y$, define $g(y)$ to be the number of distinct real solutions $x$ to the equation $f(f(x))=y$. Suppose that the set of possible values of $g(y)$ over all real numbers $y$ is exactly $\{1,5,9\}$. Compute the sum of all possible values of $f(10)$.

970

Solution: We claim that we must have $f(x)=x^{3}-3 x$. First, note that the condition $f(x)+f(-x)=0$ implies that $f$ is odd. Combined with $f$ being monic, we know that $f(x)=x^{3}+a x$ for some real number $a$. Note that $a$ must be negative; otherwise $f(x)$ and $f(f(x))$ would both be increasing and 1 would be the only possible value of $g(y)$. Now, consider the condition that the set of possible values of $g(y)$ is $\{1,5,9\}$. The fact that we can have $g(y)=9$ means that some horizontal line crosses the graph of $f(f(x)) 9$ times. Since $f(f(x))$ has degree 9 , this means that its graph will have 4 local maxima and 4 local minima. Now, suppose we start at some value of $y$ such that $g(y)=9$, and slowly increase $y$. At some point, the value of $g(y)$ will decrease. This happens when $y$ is equal to a local maximum of $f$. Since $g(y)$\\ must jump from 9 down to 5, all four local maxima must have the same value. Similarly, all four local minima must also have the same value. Since $f$ is odd, it suffices to just consider the four local maxima. The local maximum of $f(x)$ occurs when $3 x^{2}+a=0$. For convenience, let $a=-3 b^{2}$, so $f(x)=x^{3}-3 b^{2} x$. Then, the local maximum is at $x=-b$, and has a value of $f(-b)=2 b^{3}$. We consider the local maxima of $f(f(x))$ next. They occur either when $x=-b$ (meaning $f(x)$ is at a local maximum) or $f(x)=-b$. If $f(x)=-b$, then $f(f(x))=f(-b)=2 b^{3}$. Thus, we must have $f(f(-b))=f\left(2 b^{3}\right)=2 b^{3}$. This yields the equation \[ f\left(2 b^{3}\right)=8 b^{9}-3 b^{2} \cdot 2 b^{3}=2 b^{3} \] which factors as $2 b^{3}\left(b^{2}-1\right)\left(2 b^{2}+1\right)^{2}$. The only possible value of $b^{2}$ is 1 . Thus, $f(x)=x^{3}-3 x$, and our answer is $10^{3}-3 \cdot 10=970$. $\fbox{970}$.

HMMT Feb Hard

HMMT-Feb Algebra

HMMT

HMMT-Feb

team

Feb

Let $A B C$ be an acute triangle with incenter $I$ and circumcenter $O$. Assume that $\angle O I A=90^{\circ}$. Given that $A I=97$ and $B C=144$, compute the area of $\triangle A B C$.

14040

We present five different solutions and outline a sixth and seventh one. In what follows, let $a=B C$, $b=C A, c=A B$ as usual, and denote by $r$ and $R$ the inradius and circumradius. Let $s=\frac{1}{2}(a+b+c)$. In the first five solutions we will only prove that \[ \angle A I O=90^{\circ} \Longrightarrow b+c=2 a \] Let us see how this solves the problem. This lemma implies that $s=216$. If we let $E$ be the foot of $I$ on $A B$, then $A E=s-B C=72$, consequently the inradius is $r=\sqrt{97^{2}-72^{2}}=65$. Finally, the area is $s r=216 \cdot 65=14040$. First Solution. Since $O I \perp D A, A I=D I$. Now, it is a well-known fact that $D I=D B=D C$ (this is occasionally called "Fact 5"). Then by Ptolemy's Theorem, \[ D B \cdot A C+D C \cdot A B=D A \cdot B C \Longrightarrow A C+A B=2 B C \] Second Solution. As before note that $I$ is the midpoint of $A D$. Let $M$ and $N$ be the midpoints of $A B$ and $A C$, and let the reflection of $M$ across $B I$ be $P$; thus $B M=B P$. Also, $M I=P I$, but we know $M I=N I$ as $I$ lies on the circumcircle of triangle $A M N$. Consequently, we get $P I=N I$; moreover by angle chasing we have \[ \angle I N C=\angle A M I=180^{\circ}-\angle B P I=\angle I P C \] Thus triangles $I N C$ and $P I C$ are congruent ( $C I$ is a bisector) so we deduce $P C=N C$. Thus, \[ B C=B P+P C=B M+C N=\frac{1}{2}(A B+A C) \] Third Solution. We appeal to Euler's Theorem, which states that $I O^{2}=R(R-2 r)$. Thus by the Pythagorean Theorem on $\triangle A I O$ (or by Power of a Point) we may write \[ (s-a)^{2}+r^{2}=A I^{2}=R^{2}-I O^{2}=2 R r=\frac{a b c}{2 s} \] with the same notations as before. Thus, we derive that \[ \begin{aligned} a b c & =2 s\left((s-a)^{2}+r^{2}\right) \\ & =2(s-a)(s(s-a)+(s-b)(s-c)) \\ & =\frac{1}{2}(s-a)\left((b+c)^{2}-a^{2}+a^{2}-(b-c)^{2}\right) \\ & =2 b c(s-a) \end{aligned} \] From this we deduce that $2 a=b+c$, and we can proceed as in the previous solution. Fourth Solution. From Fact 5 again $(D B=D I=D C)$, drop perpendicular from $I$ to $A B$ at $E$; call $M$ the midpoint of $B C$. Then, by $A A S$ congruency on $A I E$ and $C D M$, we immediately get that $C M=A E$. As $A E=\frac{1}{2}(A B+A C-B C)$, this gives the desired conclusion. Fifth Solution. This solution avoids angle-chasing and using the fact that $B I$ and $C I$ are anglebisectors. Recall the perpendicularity lemma, where \[ W X \perp Y Z \Longleftrightarrow W Y^{2}-W Z^{2}=X Y^{2}-X Z^{2} \] Let $B^{\prime}$ be on the extension of ray $C A$ such that $A B^{\prime}=A B$. Of course, as in the proof of the angle bisector theorem, $B B^{\prime} \| A I$, meaning that $B B^{\prime} \perp I O$. Let $I^{\prime}$ be the reflection of $I$ across $A$; of course, $I^{\prime}$ is then the incenter of triangle $A B^{\prime} C^{\prime}$. Now, we have $B^{\prime} I^{2}-B I^{2}=B^{\prime} O^{2}-B O^{2}$ by the perpendicularity and by power of a point $B^{\prime} O^{2}-B O^{2}=B^{\prime} A \cdot B^{\prime} C$. Moreover $B I^{2}+B^{\prime} I^{2}=B I^{2}+B I^{\prime 2}=2 B A^{2}+2 A I^{2}$ by the median formula. Subtracting, we get $B I^{2}=A I^{2}+\frac{1}{2}(A B)(A B-A C)$. We have a similar expression for $C I$, and subtracting the two results in $B I^{2}-C I^{2}=\frac{1}{2}\left(A B^{2}-A C^{2}\right)$. Finally, \[ B I^{2}-C I^{2}=\frac{1}{4}\left[(B C+A B-A C)^{2}-(B C-A B+A C)^{2}\right] \] from which again, the result $2 B C=A B+A C$ follows. Sixth Solution, outline. Use complex numbers, setting $I=a b+b c+c a, A=-a^{2}$, etc. on the unit circle (scale the picture to fit in a unit circle; we calculate scaling factor later). Set $a=1$, and let $u=b+c$ and $v=b c$. Write every condition in terms of $u$ and $v$, and the area in terms of $u$ and $v$ too. There should be two equations relating $u$ and $v: 2 u+v+1=0$ and $u^{2}=\frac{130}{9}^{2} v$ from the right angle and the 144 to 97 ratio, respectively. The square area can be computed in terms of $u$ and $v$, because the area itself is antisymmetric so squaring it suffices. Use the first condition to homogenize (not coincidentally the factor $\left(1-b^{2}\right)\left(1-c^{2}\right)=(1+b c)^{2}-(b+c)^{2}=(1+v)^{2}-u^{2}$ from the area homogenizes perfectly. $\ldots$ because $A B \cdot A C=A I \cdot A I_{A}$, where $I_{A}$ is the $A$-excenter, and of course the way the problem is set up $A I_{A}=3 A I$.), and then we find the area of the scaled down version. To find the scaling factor simply determine $|b-c|$ by squaring it, writing in terms again of $u$ and $v$, and comparing this to the value of 144 . Seventh Solution, outline. Trigonometric solutions are also possible. One can you write everything in terms of the angles and solve the equations; for instance, the $\angle A I O=90^{\circ}$ condition can be rewritten as $\frac{1}{2} \cos \frac{B-C}{2}=2 \sin \frac{B}{2} \sin \frac{C}{2}$ and the 97 to 144 ratio condition can be rewritten as $\frac{2 \sin \frac{B}{2} 2 \sin \frac{C}{2}}{\sin A}=\frac{97}{144}$. The first equation implies $\sin \frac{A}{2}=2 \sin \frac{B}{2} \sin \frac{C}{2}$, which we can plug into the second equation to get $\cos \frac{A}{2}$. $\fbox{14040}$.

HMMT Feb Team

HMMT-Feb Team

HMMT

HMMT-Nov

guts

Nov

Let $A B$ be a segment of length 2 with midpoint $M$. Consider the circle with center $O$ and radius $r$ that is externally tangent to the circles with diameters $A M$ and $B M$ and internally tangent to the circle with diameter $A B$. Determine the value of $r$.

\frac{1}{3}

Let $X$ be the midpoint of segment $A M$. Note that $O M \perp M X$ and that $M X=\frac{1}{2}$ and $O X=\frac{1}{2}+r$ and $O M=1-r$. Therefore by the Pythagorean theorem, we have \[ O M^{2}+M X^{2}=O X^{2} \Longrightarrow(1-r)^{2}+\frac{1}{2^{2}}=\left(\frac{1}{2}+r\right)^{2} \] which we can easily solve to find that $r=\frac{1}{3}$. $\fbox{\frac{1}{3}}$.

HMMT Nov Guts

HMMT-Nov Guts

AMC

AMC8

8

N/A

Monica is tiling the floor of her 12-foot by 16-foot living room. She plans to place one-foot by one-foot square tiles to form a border along the edges of the room and to fill in the rest of the floor with two-foot by two-foot square tiles. How many tiles will she use?

87

She will place $(12\cdot2)+(14\cdot2)=52$ tiles around the border. For the inner part of the room, we have $10\cdot14=140$ square feet. Each tile takes up $4$ square feet, so he will use $\frac{140}{4}=35$ tiles for the inner part of the room. Thus, the answer is $52+35= \fbox{87}$.

AMC8 First Half

AMC8

HMMT

HMMT-Nov

thm

Nov

For dessert, Melinda eats a spherical scoop of ice cream with diameter 2 inches. She prefers to eat her ice cream in cube-like shapes, however. She has a special machine which, given a sphere placed in space, cuts it through the planes $x=n, y=n$, and $z=n$ for every integer $n$ (not necessarily positive). Melinda centers the scoop of ice cream uniformly at random inside the cube $0 \leq x, y, z \leq 1$, and then cuts it into pieces using her machine. What is the expected number of pieces she cuts the ice cream into?

7+\frac{13 \pi}{3}

Note that if we consider the division of $\mathbb{R}^{3}$ into unit cubes by the given planes, we only need to compute the sum of the probabilities that the ice cream scoop intersects each cube. There are three types of cubes that can be intersected: \begin{itemize} \item The cube $0 \leq x, y, z \leq 1$ in which the center lies, as well as the 6 face-adjacent cubes are always intersected, for a total of 7 . \item The cubes edge-adjacent to the center cube are intersected if the center of the ice cream lies within 1 unit of the connecting edge, which happens with probability $\frac{\pi}{4}$. There are 12 such cubes, for a total of $3 \pi$. \item The cubes corner-adjacent to the center cube are intersected if the center of the ice cream lies within 1 unit of the connecting corner, which happens with probability $\frac{\pi}{6}$. There are 8 such cubes, for a total of $\frac{4 \pi}{3}$. \end{itemize} Adding these all up gives our answer of $7+\frac{13 \pi}{3}$. An alternate solution is possible: We compute the number of regions into which a convex region $S$ in $\mathbb{R}^{3}$ is divided by planes: Let $a$ be the number of planes intersecting $S$. Let $b$ be the number of lines (intersections of two planes) passing through $S$. Let $c$ be the number of points (intersections of three planes) lying inside $S$. Then $S$ is divided into $a+b+c+1$ regions. Then the computation for the problem is fairly straight forward. Note that the only planes, lines, and points that can intersect the ice cream scoop $I$ are the faces, edges, and vertices of the cube $0 \leq x, y, z \leq 1$. The computation is essentially the same as in the first solution. The scoop intersects each of the 6 faces with probability 1, each of the 12 edges with probability $\frac{\pi}{4}$, and each of the 8 vertices with probability $\frac{\pi}{6}$, for a total expected number of regions $1+6+3 \pi+\frac{4 \pi}{3}=7+\frac{13 \pi}{3}$. $\fbox{7+\frac{13 \pi}{3}}$.

HMMT Nov Hard

HMMT-Nov Theme

AMC

AMC12

12A

N/A

Triangle $ABC$ has $AB=27$, $AC=26$, and $BC=25$. Let $I$ be the intersection of the internal angle bisectors of $\triangle ABC$. What is $BI$?

15

Inscribe circle $C$ of radius $r$ inside triangle $ABC$ so that it meets $AB$ at $Q$, $BC$ at $R$, and $AC$ at $S$. Note that angle bisectors of triangle $ABC$ are concurrent at the center $O$(also $I$) of circle $C$. Let $x=QB$, $y=RC$ and $z=AS$. Note that $BR=x$, $SC=y$ and $AQ=z$. Hence $x+z=27$, $x+y=25$, and $z+y=26$. Subtracting the last 2 equations we have $x-z=-1$ and adding this to the first equation we have $x=13$. By Heron's formula for the area of a triangle we have that the area of triangle $ABC$ is $\sqrt{39(14)(13)(12)}$. On the other hand the area is given by $(1/2)25r+(1/2)26r+(1/2)27r$. Then $39r=\sqrt{39(14)(13)(12)}$ so that $r^2=56$. Since the radius of circle $O$ is perpendicular to $BC$ at $R$, we have by the pythagorean theorem $BO^2=BI^2=r^2+x^2=56+169=225$ so that $BI=\fbox{15}$.

AMC12 Second Half

AMC12 A

HMMT

HMMT-Nov

guts

Nov

A real number $x$ is chosen uniformly at random from the interval $[0,1000]$. Find the probability that \[ \left\lfloor\frac{\left\lfloor\frac{x}{2.5}\right\rfloor}{2.5}\right\rfloor=\left\lfloor\frac{x}{6.25}\right\rfloor \]

\frac{9}{10}

Solution: Let $y=\frac{x}{2.5}$, so $y$ is chosen uniformly at random from $[0,400]$. Then we need \[ \left\lfloor\frac{\lfloor y\rfloor}{2.5}\right\rfloor=\left\lfloor\frac{y}{2.5}\right\rfloor \] Let $y=5 a+b$, where $0 \leq b<5$ and $a$ is an integer. Then \[ \left\lfloor\frac{\lfloor y\rfloor}{2.5}\right\rfloor=\left\lfloor\frac{5 a+\lfloor b\rfloor}{2.5}\right\rfloor=2 a+\left\lfloor\frac{\lfloor b\rfloor}{2.5}\right\rfloor \] while \[ \left\lfloor\frac{y}{2.5}\right\rfloor=\left\lfloor\frac{5 a+b}{2.5}\right\rfloor=2 a+\left\lfloor\frac{b}{2.5}\right\rfloor \] so we need $\left\lfloor\frac{\lfloor b\rfloor}{2.5}\right\rfloor=\left\lfloor\frac{b}{2.5}\right\rfloor$, where $b$ is selected uniformly at random from $[0,5]$. This can be shown to always hold except for $b \in[2.5,3)$, so the answer is $1-\frac{0.5}{5}=\frac{9}{10}$. $\fbox{\frac{9}{10}}$.

HMMT Nov Guts

HMMT-Nov Guts

HMMT

HMMT-Feb

guts

Feb

An independent set of a graph $G$ is a set of vertices of $G$ such that no two vertices among these are connected by an edge. If $G$ has 2000 vertices, and each vertex has degree 10, find the maximum possible number of independent sets that $G$ can have.

2047^{100}

The upper bound is obtained when $G$ is a disjoint union of bipartite graphs, each of which has 20 vertices with 10 in each group such that every pair of vertices not in the same group are connected. In 1991, during his study of the Cameron-Erdos conjecture on the number of sum-free sets, Noga Alon came across this problem on independent sets and conjectured that our construction gives the best bound. This problem received considerable attention due to its application to combinatorial group theory and statistical mechanics, but no solution was found until 2009, when Yufei Zhao resolved Alon's conjecture with a beautiful and elementary approach. For the reader to enjoy the full insight of Yufei's argument, we omit the proof here and refer to his paper "The Number of Independent Sets in a Regular Graph" available on his website at \href{http://web.mit.edu/yufeiz/www/indep_reg.pdf}{http://web.mit.edu/yufeiz/www/indep\_reg.pdf}. $\fbox{2047^{100}}$.

HMMT Feb Guts

HMMT-Feb Guts

HMMT

HMMT-Feb

guts

Feb

In how many ways can each square of a $4 \times 2011$ grid be colored red, blue, or yellow such that no two squares that are diagonally adjacent are the same color?

64 \cdot 3^{4020}

If we first color the board in a checkerboard pattern, it is clear that the white squares are independent of the black squares in diagonal coloring, so we calculate the number of ways to color the white squares of a $4 \times n$ board and then square it. Let $a_{n}$ be the number of ways to color the white squares of a $4 \times n$ board in this manner such that the two squares in the last column are the same color, and $b_{n}$ the number of ways to color it such that they are different. We want to find their sum $x_{n}$. We have $a_{1}=3, b_{1}=6$. Given any filled $4 \times n-1$ grid with the two white squares in the last column different, there is only 1 choice for the middle square in the $n$th row, and two choices for the outside square, 1 choice makes them the same color, 1 makes them different. If the two white squares are the same, there are 2 choices for the middle square and the outer square, so 4 choices. Of these, in 2 choices, the two new squares are the same color, and in the other 2, the two squares are different. It follows that $a_{n}=2 a_{n-1}+b_{n-1}$ and $b_{n}=2 a_{n-1}+b_{n-1}$, so $a_{n}=b_{n}$ for $n \geq 2$. We have $x_{n}=8 \cdot 3^{n-1}$ and $x_{2011}=8 \cdot 3^{2010}$. So the answer is $64 \cdot 3^{4020}$. $\fbox{64 \cdot 3^{4020}}$.

HMMT Feb Guts

HMMT-Feb Guts

AMC

AMC8

8

N/A

The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$? [asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),"$-2$"); label((0,2),"$9$"); label((2,2),"$5$"); label((2,0),"$-1$"); label((2,-2),"$8$"); label((-2,-2),"$x$"); [/asy]

8

The sum of the numbers in each row is $12$. Consider the second row. In order for the sum of the numbers in this row to equal $12$, the two shaded numbers must add up to $13$: [asy] unitsize(0.5cm); fill((-3,1)--(1,1)--(1,-1)--(-3,-1)--cycle,mediumgray); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),"$-2$"); label((0,2),"$9$"); label((2,2),"$5$"); label((2,0),"$-1$"); label((2,-2),"$8$"); label((-2,-2),"$x$"); [/asy] If two numbers add up to $13$, one of them must be at least $7$: If both shaded numbers are no more than $6$, their sum can be at most $12$. Therefore, for $x$ to be larger than the three missing numbers, $x$ must be at least $8$. We can construct a working scenario where $x=8$: [asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),"$-2$"); label((0,2),"$9$"); label((2,2),"$5$"); label((2,0),"$-1$"); label((2,-2),"$8$"); label((-2,-2),"$8$"); label((0,-2),"$-4$"); label((-2,0),"$6$"); label((0,0),"$7$"); [/asy] So, our answer is $\fbox{8}$.

AMC8 Second Half

AMC8

AMC

AMC12

12B

N/A

Two particles move along the edges of equilateral $\triangle ABC$ in the direction \[A\Rightarrow B\Rightarrow C\Rightarrow A,\] starting simultaneously and moving at the same speed. One starts at $A$, and the other starts at the midpoint of $\overline{BC}$. The midpoint of the line segment joining the two particles traces out a path that encloses a region $R$. What is the ratio of the area of $R$ to the area of $\triangle ABC$?

\frac{1}{16}

WLOG, let point $A$ be located on the origin. Let side $\overline{AB}$ be on the $x$-axis, and let point $B$ be $(0, 1)$. When one of the ants is on a vertex, the other will be on the midpoint of the opposite side. Marking these points gives us a rough view of what seem to be the vertices of a smaller equilateral triangle, that is concentric with the larger one. After observing(or imagining) the ants moving, we find that the area enclosed is indeed an equilateral triangle. Let the midpoint of $\overline{BC}$ be $D$. Then we can find the coordinates of $D$ using the equations of two lines: the extension of $\overline{AD}$ and the extension of $\overline{BC}$. The slope of any line is simply the tangent of the angle between the line and the $x$-axis. The angle between $\overline{AD}$ and the $x$-axis is $30^{\circ}$, and $tan(30) = \frac{\sqrt{3}}{3}$. Therefore the equation of the line containing $\overline{AD}$ is $y= \frac{\sqrt{3}}{3}x.$ The slope of the line containing $\overline{BC}$ is $-\sqrt{3}$, and it goes through the point $(1, 0)$. After solving for the $y$-intercept, we get $y = -\sqrt{3}x+\sqrt{3}$. Setting these equal to each other yields $D\left(\frac{3}{4}, \frac{\sqrt{3}}{4}\right).$ The midpoint of $D$ and the $A$ is the bottom-left vertex of the smaller equilateral triangle. Using the distance formula, we get the point $\left(\frac{3}{8}, \frac{\sqrt{3}}{8}\right).$ Since the smaller triangle is simply a dilation of the larger one, they both have an axis of symmetry over $x=\frac{1}{2}$. This yields the bottom-right vertex of the smaller triangle as $\left(\frac{5}{8}, \frac{\sqrt{3}}{8}\right).$ Therefore the side length of the smaller triangle is $\frac{1}{4}.$ The problem asks us for the ratio of the areas, and since area is two-dimensional, we simply square the ratio to get $\frac{1}{16} \Rightarrow \fbox{\frac{1}{16}}.$

AMC12 Final Problems

AMC12 B

AMC

AMC8

8

N/A

You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $$1.02$, with at least one coin of each type. How many dimes must you have?

1

Since you have one coin of each type, $1 + 5 + 10 + 25 = 41$ cents are already determined, leaving you with a total of $102 - 41 = 61$ cents remaining for $5$ coins. You must have $1$ more penny. If you had more than $1$ penny, you must have at least $6$ pennies to leave a multiple of $5$ for the nickels, dimes, and quarters. But you only have $5$ more coins to assign. Now you have $61 - 1 = 60$ cents remaining for $4$ coins, which may be nickels, quarters, or dimes. If you have only one more dime, that leaves $50$ cents in $3$ nickels or quarters, which is impossible. If you have two dimes, that leaves $40$ cents for $2$ nickels or quarters, which is again impossible. If you have three dimes, that leaves $30$ cents for $1$ nickel or quarter, which is still impossible. And all four remaining coins being dimes will not be enough. Therefore, you must have no more dimes to assign, and the $60$ cents in $4$ coins must be divided between the quarters and nickels. We quickly see that $2$ nickels and $2$ quarters work. Thus, the total count is $2$ quarters, $2$ nickels, $1$ penny, plus one more coin of each type that we originally subtracted. Double-checking, that gives a total $2 + 2 + 1 + 4 = 9$ coins, and a total of $2\cdot 25 + 2\cdot 5 + 1 + (1 + 5 + 10 + 25) = 102$ cents. There is only $1$ dime in that combo, so the answer is $\fbox{1}$.

AMC8 Second Half

AMC8

HMMT

HMMT-Feb

guts

Feb

After the Guts round ends, the HMMT organizers will calculate $A$, the total number of points earned over all participating teams on questions 33, 34, and 35 of this round (that is, the other estimation questions). Estimate $A$. Submit a positive integer $E$. You will receive $\max (0,25-3 \cdot|E-A|)$ points. (If you do not submit a positive integer, you will receive zero points for this question.) For your information, there are about 70 teams competing.

13

Solution: Only 8 teams scored a positive number of combined points on questions 33, 34, and 35 . A total of 3 points were scored on question 33, 6 points on question 34 , and 4 points on question 35. Extended results can be found in our archive. $\fbox{13}$.

HMMT Feb Guts

HMMT-Feb Guts

AMC

AMC10

10A

N/A

A player pays $\textdollar 5$ to play a game. A die is rolled. If the number on the die is odd, the game is lost. If the number on the die is even, the die is rolled again. In this case the player wins if the second number matches the first and loses otherwise. How much should the player win if the game is fair? (In a fair game the probability of winning times the amount won is what the player should pay.)

60

The probability of rolling an even number on the first turn is $\frac{1}{2}$ and the probability of rolling the same number on the next turn is $\frac{1}{6}$. The probability of winning is $\frac{1}{2}\cdot \frac{1}{6} =\frac{1}{12}$. If the game is to be fair, the amount paid, $5$ dollars, must be $\frac{1}{12}$ the amount of the prize money, so the answer is $\fbox{60}.$

AMC10 Second Half

AMC10 A

AMC

AMC8

8

N/A

Bill walks $\tfrac12$ mile south, then $\tfrac34$ mile east, and finally $\tfrac12$ mile south. How many miles is he, in a direct line, from his starting point?

1\tfrac14

Draw a picture. [asy] unitsize(3cm); draw((0,0)--(0,-.5)--(.75,-.5)--(.75,-1),linewidth(1pt)); label("$\frac12$",(0,0)--(0,-.5),W); label("$\frac12$",(.75,-.5)--(.75,-1),E); label("$\frac34$",(.75,-.5)--(0,-.5),N); draw((0,0)--(.75,0)--(.75,-1)--(0,-1)--cycle,gray); [/asy] Find the length of the diagonal of the rectangle to find the length of the direct line to the starting time using Pythagorean Theorem. \[\sqrt{\left(\frac12+\frac12\right)^2+\left(\frac34\right)^2} = \sqrt{1+\frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac54 = \fbox{1\tfrac14}\]

AMC8 First Half

AMC8

AMC

AMC12

12A

N/A

For each integer $n>1$, let $F(n)$ be the number of solutions to the equation $\sin{x}=\sin{(nx)}$ on the interval $[0,\pi]$. What is $\sum_{n=2}^{2007} F(n)$?

2016532

\[\sin nx - \sin x = 2\left( \cos \frac {n + 1}{2}x\right) \left( \sin \frac {n - 1}{2}x\right)\] So $\sin nx = \sin x$ if and only if $\cos \frac {n + 1}{2}x = 0$ or $\sin \frac {n - 1}{2}x = 0$. The first occurs whenever $\frac {n + 1}{2}x = (j + 1/2)\pi$, or $x = \frac {(2j + 1)\pi}{n + 1}$ for some nonnegative integer $j$. Since $x\leq \pi$, $j\leq n/2$. So there are $1 + \lfloor n/2 \rfloor$ solutions in this case. The second occurs whenever $\frac {n - 1}{2}x = k\pi$, or $x = \frac {2k\pi}{n - 1}$ for some nonnegative integer $k$. Here $k\leq \frac {n - 1}{2}$ so that there are $\left\lfloor \frac {n + 1}{2}\right\rfloor$ solutions here. However, we overcount intersections. These occur whenever \[\frac {2j + 1}{n + 1} = \frac {2k}{n - 1}\] \[k = \frac {(2j + 1)(n - 1)}{2(n + 1)}\] which is equivalent to $2(n + 1)$ dividing $(2j + 1)(n - 1)$. If $n$ is even, then $(2j + 1)(n - 1)$ is odd, so this never happens. If $n\equiv 3\pmod{4}$, then there won't be intersections either, since a multiple of 8 can't divide a number which is not even a multiple of 4. This leaves $n\equiv 1\pmod{4}$. In this case, the divisibility becomes $\frac {n + 1}{2}$ dividing $(2j + 1)\frac {n - 1}{4}$. Since $\frac {n + 1}{2}$ and $\frac {n - 1}{4}$ are relatively prime (subtracting twice the second number from the first gives 1), $\frac {n + 1}{2}$ must divide $2j + 1$. Since $j\leq \frac {n - 1}{2}$, $2j + 1\leq n < 2\cdot \frac {n + 1}{2}$. Then there is only one intersection, namely when $j = \frac {n - 1}{4}$. Therefore we find $F(n)$ is equal to $1 + \lfloor n/2 \rfloor + \left \lfloor \frac {n + 1}{2}\right\rfloor = n + 1$, unless $n\equiv 1\pmod{4}$, in which case it is one less, or $n$. The problem may then be finished as in Solution 1. Note from Williamgolly: An easier way to see that there is an extra point of intersection at $n \equiv 1 \pmod{4}$ is that $v_2(2j+1)=0,$ so we must have $v_2(n-1)>v_2(n+1)$ since $v_2(2k) \geq 1 >v_2(2j+1).$ Therefore, we suspect that $n \equiv 1 \pmod{4}$ has an extra point of intersection. Testing, we see this is true. $\fbox{2016532}$.

AMC12 Final Problems

AMC12 A

AMC

AMC12

12A

N/A

Jerry starts at $0$ on the real number line. He tosses a fair coin $8$ times. When he gets heads, he moves $1$ unit in the positive direction; when he gets tails, he moves $1$ unit in the negative direction. The probability that he reaches $4$ at some time during this process $\frac{a}{b},$ where $a$ and $b$ are relatively prime positive integers. What is $a + b?$ (For example, he succeeds if his sequence of tosses is $HTHHHHHH.$)

151

For $6$ to $8$ heads, we are guaranteed to hit $4$ heads, so the sum here is $\binom{8}{2}+\binom{8}{1}+\binom{8}{0}=28+8+1=37$. For $4$ heads, you have to hit the $4$ heads at the start so there's only one way, $1$. For $5$ heads, we either start off with $4$ heads, which gives us $_4\text{C}_1=4$ ways to arrange the other flips, or we start off with five heads and one tail, which has $6$ ways minus the $2$ overlapping cases, $\text{HHHHHTTT}$ and $\text{HHHHTHTT}$. Total ways: $8$. Then we sum to get $46$. There are a total of $2^8=256$ possible sequences of $8$ coin flips, so the probability is $\frac{46}{256}=\frac{23}{128}$. Summing, we get $23+128=\fbox{151}$.

AMC12 Second Half

AMC12 A

AMC

AMC12

12B

N/A

Let $ABCD$ be a convex quadrilateral with $BC=2$ and $CD=6.$ Suppose that the centroids of $\triangle ABC,\triangle BCD,$ and $\triangle ACD$ form the vertices of an equilateral triangle. What is the maximum possible value of the area of $ABCD$?

12+10\sqrt3

Place an origin at $A$, and assign position vectors of $B = \vec{p}$ and $D = \vec{q}$. Since $AB$ is not parallel to $AD$, vectors $\vec{p}$ and $\vec{q}$ are linearly independent, so we can write $C = m\vec{p} + n\vec{q}$ for some constants $m$ and $n$. Now, recall that the centroid of a triangle $\triangle XYZ$ has position vector $\frac{1}{3}\left(\vec{x}+\vec{y}+\vec{z}\right)$. Thus the centroid of $\triangle ABC$ is $g_1 = \frac{1}{3}(m+1)\vec{p} + \frac{1}{3}n\vec{q}$; the centroid of $\triangle BCD$ is $g_2 = \frac{1}{3}(m+1)\vec{p} + \frac{1}{3}(n+1)\vec{q}$; and the centroid of $\triangle ACD$ is $g_3 = \frac{1}{3}m\vec{p} + \frac{1}{3}(n+1)\vec{q}$. Hence $\overrightarrow{G_{1}G_{2}} = \frac{1}{3}\vec{q}$, $\overrightarrow{G_{2}G_{3}} = -\frac{1}{3}\vec{p}$, and $\overrightarrow{G_{3}G_{1}} = \frac{1}{3}\vec{p} - \frac{1}{3}\vec{q}$. For $\triangle G_{1}G_{2}G_{3}$ to be equilateral, we need $\left|\overrightarrow{G_{1}G_{2}}\right| = \left|\overrightarrow{G_{2}G_{3}}\right| \Rightarrow \left|\vec{p}\right| = \left|\vec{q}\right| B = AD$. Further, $\left|\overrightarrow{G_{1}G_{2}}\right| = \left|\overrightarrow{G_{1}G_{3}}\right| \Rightarrow \left|\vec{p}\right| = \left|\vec{p} - \vec{q}\right| = BD$. Hence we have $AB = AD = BD$, so $\triangle ABD$ is equilateral. Now let the side length of $\triangle ABD$ be $k$, and let $\angle BCD = \theta$. By the Law of Cosines in $\triangle BCD$, we have $k^2 = 2^2 + 6^2 - 2 \cdot 2 \cdot 6 \cdot \cos{\theta} = 40 - 24\cos{\theta}$. Since $\triangle ABD$ is equilateral, its area is $\frac{\sqrt{3}}{4}k^2 = 10\sqrt{3} - 6\sqrt{3}\cos{\theta}$, while the area of $\triangle BCD$ is $\frac{1}{2} \cdot 2 \cdot 6 \cdot \sin{\theta} = 6 \sin{\theta}$. Thus the total area of $ABCD$ is $10\sqrt{3} + 6\left(\sin{\theta} - \sqrt{3}\cos{\theta}\right) = 10\sqrt{3} + 12\left(\frac{1}{2} \sin{\theta} - \frac{\sqrt{3}}{2}\cos{\theta}\right) = 10\sqrt{3}+12\sin{\left(\theta-60^{\circ}\right)}$, where in the last step we used the subtraction formula for $\sin$. Alternatively, we can use calculus to find the local maximum. Observe that $\sin{\left(\theta-60^{\circ}\right)}$ has maximum value $1$ when e.g. $\theta = 150^{\circ}$, which is a valid configuration, so the maximum area is $10\sqrt{3} + 12(1) = \fbox{12+10\sqrt3}$.

AMC12 Final Problems

AMC12 B

AMC

AMC10

10A

N/A

In how many ways can the sequence $1,2,3,4,5$ be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?

32

We write out the $5!=120$ cases, then filter out the valid ones: $13254,14253,14352,15243,15342,21435,21534,23154,24153,24351,25143,25341,\linebreak 31425,31524,32415,32514,34152,34251,35142,35241,41325,41523,42315,42513,\linebreak 43512,45132,45231,51324,51423,52314,52413,53412.$ We count these out and get $\fbox{32}$ permutations that work.

AMC10 Second Half

AMC10 A

AMC

AMC8

8

N/A

Twelve friends met for dinner at Oscar's Overstuffed Oyster House, and each ordered one meal. The portions were so large, there was enough food for $18$ people. If they shared, how many meals should they have ordered to have just enough food for the $12$ of them?

8

Set up the proportion $\frac{12\ \text{meals}}{18\ \text{people}}=\frac{x\ \text{meals}}{12\ \text{people}}$. Solving for $x$ gives us $x= \fbox{8}$.

AMC8 First Half

AMC8

AMC

AMC8

8

N/A

For any positive integer $M$, the notation $M!$ denotes the product of the integers $1$ through $M$. What is the largest integer $n$ for which $5^n$ is a factor of the sum $98!+99!+100!$ ?

26

Factoring out $98!+99!+100!$, we have $98! (1+99+99*100)$, which is $98! (10000)$. Next, $98!$ has $\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22$ factors of $5$. The $19$ is because of all the multiples of $5$.The $3$ is because of all the multiples of $25$. Now, $10,000$ has $4$ factors of $5$, so there are a total of $22 + 4 = \fbox{26}$ factors of $5$.

AMC8 Second Half

AMC8

HMMT

HMMT-Feb

geo

Feb

Let $A, B, C, D$ be four points on a circle in that order. Also, $A B=3, B C=5, C D=6$, and $D A=4$. Let diagonals $A C$ and $B D$ intersect at $P$. Compute $\frac{A P}{C P}$.

\frac{2}{5}

Note that $\triangle A P B \sim \triangle D P C$ so $\frac{A P}{A B}=\frac{D P}{C D}$. Similarly, $\triangle B P C \sim \triangle A P D$ so $\frac{C P}{B C}=\frac{D P}{D A}$. Dividing these two equations yields \[ \frac{A P}{C P}=\frac{A B \cdot D A}{B C \cdot C D}=\frac{2}{5} \] $\fbox{\frac{2}{5}}$.

HMMT Feb Easy

HMMT-Feb Geometry

HMMT

HMMT-Nov

thm

Nov

Sandy likes to eat waffles for breakfast. To make them, she centers a circle of waffle batter of radius $3 \mathrm{~cm}$ at the origin of the coordinate plane and her waffle iron imprints non-overlapping unit-square holes centered at each lattice point. How many of these holes are contained entirely within the area of the waffle?

21

First, note that each divet must have its sides parallel to the coordinate axes; if the divet centered at the lattice point $(a, b)$ does not have this orientation, then it contains the point $(a+1 / 2, b)$ in its interior, so it necessarily overlaps with the divet centered at $(a+1, b)$. If we restrict our attention to one quadrant, we see geometrically that the divets centered at $(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0)$, and $(2,1)$ are completely contained in the waffle, and no others are. We can make this more rigorous by considering the set of points $(x, y)$ such that $x^{2}+y^{2}<9$. We count 1 divet centered at the origin, 8 divets centered on the axes that are not centered at the origin, and 12 divets not centered on the axes, for a total of 21 divets. $\fbox{21}$.

HMMT Nov Easy

HMMT-Nov Theme

HMMT

HMMT-Nov

gen

Nov

Marisa has a collection of $2^{8}-1=255$ distinct nonempty subsets of $\{1,2,3,4,5,6,7,8\}$. For each step she takes two subsets chosen uniformly at random from the collection, and replaces them with either their union or their intersection, chosen randomly with equal probability. (The collection is allowed to contain repeated sets.) She repeats this process $2^{8}-2=254$ times until there is only one set left in the collection. What is the expected size of this set?

\frac{1024}{255}

It suffices to compute the probability of each number appearing in the final subset. For any given integer $n \in[1,8]$, there are $2^{7}=128$ subsets with $n$ and $2^{7}-1=127$ without. When we focus on only this element, each operation is equivalent to taking two random sets and discarding one of them randomly. Therefore there is a $\frac{128}{255}$ probability that $n$ is in the final subset, and the expected value of its size is $8 \cdot \frac{128}{255}=\frac{1024}{255}$. Alternatively, since $|A|+|B|=|A \cup B|+|A \cap B|$, the expected value of the average size of all remaining subsets at a given step is constant, so the answer is simply the average size of all 255 subsets, which is $\frac{8 \cdot 128}{255}=\frac{1024}{255}$. $\fbox{\frac{1024}{255}}$.

HMMT Nov Hard

HMMT-Nov General

HMMT

HMMT-Nov

thm

Nov

Alice is once again very bored in class. On a whim, she chooses three primes $p, q, r$ independently and uniformly at random from the set of primes of at most 30. She then calculates the roots of $p x^{2}+q x+r$. What is the probability that at least one of her roots is an integer?

\frac{3}{200}

Solution: Since all of the coefficients are positive, any root $x$ must be negative. Moreover, by the rational root theorem, in order for $x$ to be an integer we must have either $x=-1$ or $x=-r$. So we must have either $p r^{2}-q r+r=0 \Longleftrightarrow p r=q-1$ or $p-q+r=0$. Neither of these cases are possible if all three primes are odd, so we know so we know that one of the primes is even, hence equal to 2. After this we can do a casework check; the valid triples of $(p, q, r)$ are $(2,5,3),(2,7,5),(2,13,11)$, $(2,19,17),(2,5,2),(2,7,3),(2,11,5),(2,23,11)$, allowing for $p$ and $r$ to be swapped. This leads to 15 valid triples out of 1000 (there are 10 primes less than 30 ). $\fbox{\frac{3}{200}}$.

HMMT Nov Easy

HMMT-Nov Theme

HMMT

HMMT-Nov

guts

Nov

Let $A B C D$ be a rectangle with $A B=3$ and $B C=7$. Let $W$ be a point on segment $A B$ such that $A W=1$. Let $X, Y, Z$ be points on segments $B C, C D, D A$, respectively, so that quadrilateral $W X Y Z$ is a rectangle, and $B X<X C$. Determine the length of segment $B X$.

\frac{7-\sqrt{41}}{2}

We note that \[ \angle Y X C=90-\angle W X B=\angle X W B=90-\angle A W Z=\angle A Z W \] gives us that $X Y C \cong Z W A$ and $X Y Z \sim W X B$. Consequently, we get that $Y C=A W=1$. From $X Y Z \sim W X B$, we get that \[ \frac{B X}{B W}=\frac{C Y}{C X} \Rightarrow \frac{B X}{2}=\frac{1}{7-B X} \] from which we get \[ B X^{2}-7 B X+2=0 \Rightarrow B X=\frac{7-\sqrt{41}}{2} \] (since we have $B X<C X$ ). Guts Round $\fbox{\frac{7-\sqrt{41}}{2}}$.

HMMT Nov Guts

HMMT-Nov Guts

HMMT

HMMT-Feb

comb

Feb

Alice Czarina is bored and is playing a game with a pile of rocks. The pile initially contains 2015 rocks. At each round, if the pile has $N$ rocks, she removes $k$ of them, where $1 \leq k \leq N$, with each possible $k$ having equal probability. Alice Czarina continues until there are no more rocks in the pile. Let $p$ be the probability that the number of rocks left in the pile after each round is a multiple of 5 . If $p$ is of the form $5^{a} \cdot 31^{b} \cdot \frac{c}{d}$, where $a, b$ are integers and $c, d$ are positive integers relatively prime to $5 \cdot 31$, find $a+b$.

-501

We claim that $p=\frac{1}{5} \frac{6}{10} \frac{11}{15} \frac{16}{20} \cdots \frac{2006}{2010} \frac{2011}{2015}$. Let $p_{n}$ be the probability that, starting with $n$ rocks, the number of rocks left after each round is a multiple of 5. Indeed, using recursions we have \[ p_{5 k}=\frac{p_{5 k-5}+p_{5 k-10}+\cdots+p_{5}+p_{0}}{5 k} \] for $k \geq 1$. For $k \geq 2$ we replace $k$ with $k-1$, giving us \[ \begin{gathered} p_{5 k-5}=\frac{p_{5 k-10}+p_{5 k-15}+\cdots+p_{5}+p_{0}}{5 k-5} \\ \Longrightarrow(5 k-5) p_{5 k-5}=p_{5 k-10}+p_{5 k-15}+\cdots+p_{5}+p_{0} \end{gathered} \] Substituting this back into the first equation, we have \[ 5 k p_{5 k}=p_{5 k-5}+\left(p_{5 k-10}+p_{5 k-15}+\cdots+p_{5}+p_{0}\right)=p_{5 k-5}+(5 k-5) p_{5 k-5} \] which gives $p_{5 k}=\frac{5 k-4}{5 k} p_{5 k-5}$. Using this equation repeatedly along with the fact that $p_{0}=1$ proves the claim. Now, the power of 5 in the denominator is $v_{5}(2015 !)=403+80+16+3=502$, and 5 does not divide any term in the numerator. Hence $a=-502$. (The sum counts multiples of 5 plus multiples of $5^{2}$ plus multiples of $5^{3}$ and so on; a multiple of $5^{n}$ but not $5^{n+1}$ is counted exactly $n$ times, as desired.) Noting that $2015=31 \cdot 65$, we found that the numbers divisible by 31 in the numerator are those of the form $31+155 k$ where $0 \leq k \leq 12$, including $31^{2}=961$; in the denominator they are of the form $155 k$ where $1 \leq k \leq 13$. Hence $b=(13+1)-13=1$ where the extra 1 comes from $31^{2}$ in the numerator. Thus $a+b=-501$. $\fbox{-501}$.

HMMT Feb Easy

HMMT-Feb Combinatorics

HMMT

HMMT-Feb

guts

Feb

Let $t=2016$ and $p=\ln 2$. Evaluate in closed form the sum \[ \sum_{k=1}^{\infty}\left(1-\sum_{n=0}^{k-1} \frac{e^{-t} t^{n}}{n !}\right)(1-p)^{k-1} p \]

1-\left(\frac{1}{2}\right)^{2016}

Let $q=1-p$. Then \[ \begin{aligned} \sum_{k=1}^{\infty}\left(1-\sum_{n=0}^{k-1} \frac{e^{-t} t^{n}}{n !}\right) q^{k-1} p & =\sum_{k=1}^{\infty} q^{k-1} p-\sum_{k=1}^{\infty} \sum_{n=0}^{k-1} \frac{e^{-t} t^{n}}{n !} q^{k-1} p \\ & =1-\sum_{k=1}^{\infty} \sum_{n=0}^{k-1} \frac{e^{-t} t^{n}}{n !} q^{k-1} p \\ & =1-\sum_{n=0}^{\infty} \sum_{k=n+1}^{\infty} \frac{e^{-t} t^{n}}{n !} q^{k-1} p \\ & =1-\sum_{n=0}^{\infty} \frac{e^{-t} t^{n}}{n !} q^{n} \\ & =1-\sum_{n=0}^{\infty} \frac{e^{-t}(q t)^{n}}{n !}=1-e^{-t} e^{q t}=1-e^{-p t} \end{aligned} \] Thus the answer is $1-\left(\frac{1}{2}\right)^{2016}$. $\fbox{1-\left(\frac{1}{2}\right)^{2016}}$.

HMMT Feb Guts

HMMT-Feb Guts

HMMT

HMMT-Feb

guts

Feb

Let $A B C$ be a triangle with $A B=3, B C=4$, and $C A=5$. Let $A_{1}, A_{2}$ be points on side $B C$, $B_{1}, B_{2}$ be points on side $C A$, and $C_{1}, C_{2}$ be points on side $A B$. Suppose that there exists a point $P$ such that $P A_{1} A_{2}, P B_{1} B_{2}$, and $P C_{1} C_{2}$ are congruent equilateral triangles. Find the area of convex hexagon $A_{1} A_{2} B_{1} B_{2} C_{1} C_{2}$.

\frac{12+22 \sqrt{3}}{15}

Since $P$ is the shared vertex between the three equilateral triangles, we note that $P$ is the incenter of $A B C$ since it is equidistant to all three sides. Since the area is 6 and the semiperimeter is also 6 , we can calculate the inradius, i.e. the altitude, as 1, which in turn implies that the side length of the equilateral triangle is $\frac{2}{\sqrt{3}}$. Furthermore, since the incenter is the intersection of angle bisectors, it is easy to see that $A B_{2}=A C_{1}, B C_{2}=B A_{1}$, and $C A_{2}=C B_{1}$. Using the fact that the altitudes from $P$ to $A B$ and $C B$ form a square with the sides, we use the side lengths of the equilateral triangle to compute that $A B_{2}=A C_{1}=2-\frac{1}{\sqrt{3}}, B A_{1}=B C_{2}=1-\frac{1}{\sqrt{3}}$, and $C B_{1}=C A_{2}=3-\frac{1}{\sqrt{3}}$. We have that the area of the hexagon is therefore \[ 6-\left(\frac{1}{2}\left(2-\frac{1}{\sqrt{3}}\right)^{2} \cdot \frac{4}{5}+\frac{1}{2}\left(1-\frac{1}{\sqrt{3}}\right)^{2}+\frac{1}{2}\left(3-\frac{1}{\sqrt{3}}\right)^{2} \cdot \frac{3}{5}\right)=\frac{12+22 \sqrt{3}}{15} \] $\fbox{\frac{12+22 \sqrt{3}}{15}}$.

HMMT Feb Guts

HMMT-Feb Guts

HMMT

HMMT-Nov

gen

Nov

Consider triangle $A B C$ where $B C=7, C A=8$, and $A B=9 . D$ and $E$ are the midpoints of $B C$ and $C A$, respectively, and $A D$ and $B E$ meet at $G$. The reflection of $G$ across $D$ is $G^{\prime}$, and $G^{\prime} E$ meets $C G$ at $P$. Find the length $P G$.

\frac{\sqrt{145}}{9}

Observe that since $G^{\prime}$ is a reflection and $G D=\frac{1}{2} A G$, we have $A G=G G^{\prime}$ and therefore, $P$ is the centroid of triangle $A C G^{\prime}$. Thus, extending $C G$ to hit $A B$ at $F, P G=\frac{1}{3} C G=$ $\frac{2}{9} C F=\frac{2}{9} \sqrt{\frac{2\left(8^{2}+7^{2}\right)-9^{2}}{4}}=\frac{\sqrt{145}}{9}$ by the formula for the length of a median. $\fbox{\frac{\sqrt{145}}{9}}$.

HMMT Nov Hard

HMMT-Nov General

HMMT

HMMT-Nov

guts

Nov

A circle $\omega_{1}$ of radius 15 intersects a circle $\omega_{2}$ of radius 13 at points $P$ and $Q$. Point $A$ is on line $P Q$ such that $P$ is between $A$ and $Q . R$ and $S$ are the points of tangency from $A$ to $\omega_{1}$ and $\omega_{2}$, respectively, such that the line $A S$ does not intersect $\omega_{1}$ and the line $A R$ does not intersect $\omega_{2}$. If $P Q=24$ and $\angle R A S$ has a measure of $90^{\circ}$, compute the length of $A R$.

\sqrt{14+\sqrt{97}}

Let $O_{1}$ be the center of $\omega_{1}$ and $O_{2}$ be the center of $\omega_{2}$. Then $O_{1} O_{2}$ and $P Q$ are perpendicular. Let their point of intersection be $X$. Using the Pythagorean theorem, the fact that $P Q=24$, and our knowledge of the radii of the circles, we can compute that $O_{1} X=9$ and $O_{2} X=5$, so $O_{1} O_{2}=14$. Let $S O_{1}$ and $R O_{2}$ meet at $Y$. Then $S A R Y$ is a square, say of side length $s$. Then $O_{1} Y=s-15$ and $O_{2} Y=s-13$. So, $O_{1} O_{2} Y$ is a right triangle with sides $14, s-15$, and $s-13$. By the Pythagorean theorem, $(s-13)^{2}+(s-15)^{2}=14^{2}$. We can write this as $2 s^{2}-4 \cdot 14 s+198=0$, or $s^{2}-28 s+99=0$. The quadratic formula then gives $s=\frac{28 \pm \sqrt{388}}{2}=14 \pm \sqrt{97}$. Since $14-\sqrt{97}<15$ and $Y O_{1}>15$, we can discard the root of $14-\sqrt{97}$, and the answer is therefore $14+\sqrt{97}$. $\fbox{\sqrt{14+\sqrt{97}}}$.

HMMT Nov Guts

HMMT-Nov Guts

AMC

AMC10

10A

N/A

Many Gothic cathedrals have windows with portions containing a ring of congruent circles that are circumscribed by a larger circle, In the figure shown, the number of smaller circles is four. What is the ratio of the sum of the areas of the four smaller circles to the area of the larger circle? [asy] unitsize(6mm); defaultpen(linewidth(.8pt)); draw(Circle((0,0),1+sqrt(2))); draw(Circle((sqrt(2),0),1)); draw(Circle((0,sqrt(2)),1)); draw(Circle((-sqrt(2),0),1)); draw(Circle((0,-sqrt(2)),1)); [/asy]

4(3-2\sqrt2)

Draw some of the radii of the small circles as in the picture below. [asy] unitsize(12mm); defaultpen(linewidth(.8pt)); draw(Circle((0,0),1+sqrt(2))); draw(Circle((sqrt(2),0),1)); draw(Circle((0,sqrt(2)),1)); draw(Circle((-sqrt(2),0),1)); draw(Circle((0,-sqrt(2)),1)); draw( (sqrt(2),0) -- (0,sqrt(2)) -- (-sqrt(2),0) -- (0,-sqrt(2)) -- cycle ); draw( (0,sqrt(2)) -- (0,1+sqrt(2)) ); draw( (0,-sqrt(2)) -- (0,-1-sqrt(2)) ); draw( (0,sqrt(2)) -- (0,-sqrt(2)), dashed ); [/asy] Out of symmetry, the quadrilateral in the center must be a square. Its side is $2r$, and therefore its diagonal is $2r\sqrt{2}$. We can now compute the length of the vertical diameter of the large circle as $2r + 2r\sqrt{2}$. Hence $2R=2r + 2r\sqrt{2}$, and thus $R=r+r\sqrt{2}=r(1+\sqrt{2})$. Then the area of the large circle is $L = \pi R^2 = \pi r^2 (1+\sqrt 2)^2 = \pi r^2 (3+2\sqrt 2)$. The area of four small circles is $S = 4\pi r^2$. Hence their ratio is: \begin{align} \frac SL & = \frac{4\pi r^2}{\pi r^2 (3+2\sqrt 2)} \\ & = \frac 4{3+2\sqrt 2} \\ & = \frac 4{3+2\sqrt 2} \cdot \frac{3-2\sqrt 2}{3 - 2\sqrt 2} \\ & = \frac{4(3 - 2\sqrt 2)}{3^2 - (2\sqrt 2)^2} \\ & = \frac{4(3 - 2\sqrt 2)}1 \\ & = \fbox{4(3-2\sqrt2)} \end{align}

AMC10 Final Problems

AMC10 A

HMMT

HMMT-Feb

guts

Feb

Find the smallest possible value of $x+y$ where $x, y \geq 1$ and $x$ and $y$ are integers that satisfy $x^{2}-29 y^{2}=1$

11621

Continued fraction convergents to $\sqrt{29}$ are $5, \frac{11}{2}, \frac{16}{3}, \frac{27}{5}, \frac{70}{13}$ and you get $70^{2}-29 \cdot 13^{2}=-1$ so since $(70+13 \sqrt{29})^{2}=9801+1820 \sqrt{29}$ the answer is $9801+1820=11621$ $\fbox{11621}$.

HMMT Feb Guts

HMMT-Feb Guts

HMMT

HMMT-Feb

guts

Feb

A contest has six problems worth seven points each. On any given problem, a contestant can score either 0,1 , or 7 points. How many possible total scores can a contestant achieve over all six problems?

28

For $0 \leq k \leq 6$, to obtain a score that is $k(\bmod 6)$ exactly $k$ problems must get a score of 1 . The remaining $6-k$ problems can generate any multiple of 7 from 0 to $7(6-k)$, of which there are $7-k$. So the total number of possible scores is $\sum_{k=0}^{6}(7-k)=28$. $\fbox{28}$.

HMMT Feb Guts

HMMT-Feb Guts

HMMT

HMMT-Feb

guts

Feb

Let $A B C$ be an acute triangle with $A$-excircle $\Gamma$. Let the line through $A$ perpendicular to $B C$ intersect $B C$ at $D$ and intersect $\Gamma$ at $E$ and $F$. Suppose that $A D=D E=E F$. If the maximum value of $\sin B$ can be expressed as $\frac{\sqrt{a}+\sqrt{b}}{c}$ for positive integers $a, b$, and $c$, compute the minimum possible value of $a+b+c$.

705

Solution: First note that we can assume $A B<A C$. Suppose $\Gamma$ is tangent to $B C$ at $T$. Let $A D=$ $D E=E F=x$. Then, by Power of a Point, we have $D T^{2}=D E \cdot D F=x \cdot 2 x=2 x^{2} \Longrightarrow D T=x \sqrt{2}$. Note that $C T=s-b$, and since the length of the tangent from $A$ to $\Gamma$ is $s$, we have $s^{2}=A E \cdot A F=6 x^{2}$, so $C T=x \sqrt{6}-b$. Since $B C=B D+D T+T C$, we have $B D=B C-x \sqrt{2}-(x \sqrt{6}-b)=a+b-x(\sqrt{2}+\sqrt{6})$. Since $a+b=2 s-c=2 x \sqrt{6}-c$, we have $B D=x(\sqrt{6}-\sqrt{2})-c$. Now, by Pythagorean Theorem, we have $c^{2}=A B^{2}=A D^{2}+B D^{2}=x^{2}+[x(\sqrt{6}-\sqrt{2})-c]^{2}$. Simplifying gives $x^{2}(9-4 \sqrt{3})=x c(2 \sqrt{6}-2 \sqrt{2})$. This yields \[ \frac{x}{c}=\frac{2 \sqrt{6}-2 \sqrt{2}}{9-4 \sqrt{3}}=\frac{6 \sqrt{2}+10 \sqrt{6}}{33}=\frac{\sqrt{72}+\sqrt{600}}{33} \] $\fbox{705}$.

HMMT Feb Guts

HMMT-Feb Guts

HMMT

HMMT-Nov

guts

Nov

Let $A B C$ be a triangle with $A B=5, B C=8, C A=11$. The incircle $\omega$ and $A$-excircle ${ }^{1} \Gamma$ are centered at $I_{1}$ and $I_{2}$, respectively, and are tangent to $B C$ at $D_{1}$ and $D_{2}$, respectively. Find the ratio of the area of $\triangle A I_{1} D_{1}$ to the area of $\triangle A I_{2} D_{2}$.

\frac{1}{9}

Let $D_{1}^{\prime}$ and $D_{2}^{\prime}$ be the points diametrically opposite $D_{1}$ and $D_{2}$ on the incircle and $A$-excircle, respectively. As $I_{x}$ is the midpoint of $D_{x}$ and $D_{x}^{\prime}$, we have \[ \frac{\left[A I_{1} D_{1}\right]}{\left[A I_{2} D_{2}\right]}=\frac{\left[A D_{1} D_{1}^{\prime}\right]}{\left[A D_{2} D_{2}^{\prime}\right]} \] Now, $\triangle A D_{1} D_{1}^{\prime}$ and $\triangle A D_{2} D_{2}^{\prime}$ are homothetic with ratio $\frac{r}{r_{A}}=\frac{s-a}{s}$, where $r$ is the inradius, $r_{A}$ is the $A$-exradius, and $s$ is the semiperimeter. Our answer is thus \[ \left(\frac{s-a}{s}\right)^{2}=\left(\frac{4}{12}\right)=\frac{1}{9} \] $\fbox{\frac{1}{9}}$.

HMMT Nov Guts

HMMT-Nov Guts

HMMT

HMMT-Feb

alg

Feb

Let $\otimes$ be a binary operation that takes two positive real numbers and returns a positive real number. Suppose further that $\otimes$ is continuous, commutative $(a \otimes b=b \otimes a)$, distributive across multiplication $(a \otimes(b c)=(a \otimes b)(a \otimes c))$, and that $2 \otimes 2=4$. Solve the equation $x \otimes y=x$ for $y$ in terms of $x$ for $x>1$.

y=\sqrt{2}

We note that $\left(a \otimes b^{k}\right)=(a \otimes b)^{k}$ for all positive integers $k$. Then for all rational numbers $\frac{p}{q}$ we have $a \otimes b^{\frac{p}{q}}=\left(a \otimes b^{\frac{1}{q}}\right)^{p}=(a \otimes b)^{\frac{p}{q}}$. So by continuity, for all real numbers $a, b$, it follows that $2^{a} \otimes 2^{b}=(2 \otimes 2)^{a b}=4^{a b}$. Therefore given positive reals $x, y$, we have $x \otimes y=2^{\log _{2}(x)} \otimes 2^{\log _{2}(y)}=$ $4^{\log _{2}(x) \log _{2}(y)}$. If $x=4^{\log _{2}(x) \log _{2}(y)}=2^{2 \log _{2}(x) \log _{2}(y)}$ then $\log _{2}(x)=2 \log _{2}(x) \log _{2}(y)$ and $1=2 \log _{2}(y)=\log _{2}\left(y^{2}\right)$. Thus $y=\sqrt{2}$ regardless of $x$. $\fbox{y=\sqrt{2}}$.

HMMT Feb Hard

HMMT-Feb Algebra

AMC

AMC12

12A

N/A

A circle with center $C$ is tangent to the positive $x$ and $y$-axes and externally tangent to the circle centered at $(3,0)$ with radius $1$. What is the sum of all possible radii of the circle with center $C$?

8

Let $r$ be the radius of our circle. For it to be tangent to the positive $x$ and $y$ axes, we must have $C=(r,r)$. For the circle to be externally tangent to the circle centered at $(3,0)$ with radius $1$, the distance between $C$ and $(3,0)$ must be exactly $r+1$. By the Pythagorean theorem the distance between $(r,r)$ and $(3,0)$ is $\sqrt{ (r-3)^2 + r^2 }$, hence we get the equation $(r-3)^2 + r^2 = (r+1)^2$. Simplifying, we obtain $r^2 - 8r + 8 = 0$. By Vieta's formulas the sum of the two roots of this equation is $\fbox{8}$. (We should actually solve for $r$ to verify that there are two distinct positive roots. In this case we get $r=4\pm 2\sqrt 2$. This is generally a good rule of thumb, but is not necessary as all of the available answers are integers, and the equation obviously doesn't factor as integers. You can also tell that there are two positive roots based on the visual interpretation.) [asy] unitsize(0.5cm); defaultpen(0.8); filldraw( Circle( (3,0), 1 ), lightgray, black ); draw( (0,0) -- (15,0), Arrow ); draw( (0,0) -- (0,15), Arrow ); draw( (0,0) -- (15,15), dashed ); real r1 = 4 - 2*sqrt(2), r2 = 4 + 2*sqrt(2); pair S1=(r1,r1), S2=(r2,r2); dot(S1); dot(S2); dot((3,0)); draw( Circle(S1,r1) ); draw( Circle(S2,r2) ); [/asy]

AMC12 Second Half

AMC12 A

HMMT

HMMT-Feb

alg

Feb

Let $\mathbb{N}$ be the set of positive integers, and let $f: \mathbb{N} \rightarrow \mathbb{N}$ be a function satisfying \begin{itemize} $f(1)=1$ for $n \in \mathbb{N}, f(2 n)=2 f(n)$ and $f(2 n+1)=2 f(n)-1$. \end{itemize} Determine the sum of all positive integer solutions to $f(x)=19$ that do not exceed 2019.

1889

For $n=2^{a_{0}}+2^{a_{1}}+\cdots+2^{a_{k}}$ where $a_{0}>a_{1}>\cdots>a_{k}$, we can show that $f(n)=2^{a_{0}}-2^{a_{1}}-\cdots-2^{a_{k}}=$ $2^{a_{0}+1}-n$ by induction: the base case $f(1)=1$ clearly holds; for the inductive step, when $n$ is even\\ we note that $f(n)=2 f\left(\frac{n}{2}\right)=2\left(2^{a_{0}}-\frac{n}{2}\right)=2^{a_{0}+1}-n$ as desired, and when $n$ is odd we also have $f(n)=2 f\left(\frac{n-1}{2}\right)-1=2\left(2^{a_{0}}-\frac{n-1}{2}\right)-1=2^{a_{0}+1}-n$, again as desired. Since $19=f(n) \leq 2^{a_{0}} \leq n$, we have $a_{0} \geq 5$ and $n=2^{a_{0}+1}-19 \leq 2019$ gives $a_{0} \leq 9$. So the answer is $\sum_{a=5}^{9}\left(2^{a+1}-19\right)=\left(2^{11}-2^{6}\right)-19 \cdot 5=1889$. $\fbox{1889}$.

HMMT Feb Easy

HMMT-Feb Algebra

AMC

AMC10

10B

N/A

In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible? [asy] size(110); pair A, B, C, D, E, F; A = (0,0); B = (1,0); C = (2,0); D = rotate(60, A)*B; E = B + D; F = rotate(60, A)*C; draw(Circle(A, 0.5)); draw(Circle(B, 0.5)); draw(Circle(C, 0.5)); draw(Circle(D, 0.5)); draw(Circle(E, 0.5)); draw(Circle(F, 0.5)); [/asy]

12

First we figure out the number of ways to put the $3$ blue disks. Denote the spots to put the disks as $1-6$ from left to right, top to bottom. The cases to put the blue disks are $(1,2,3),(1,2,4),(1,2,5),(1,2,6),(2,3,5),(1,4,6)$. For each of those cases we can easily figure out the number of ways for each case, so the total amount is $2+2+3+3+1+1 = \fbox{12}$.

AMC10 Second Half

AMC10 B

AMC

AMC10

10B

N/A

A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of $34$ points, and the Cougars won by a margin of $14$ points. How many points did the Panthers score?

10

Let $x$ be the number of points scored by the Cougars, and $y$ be the number of points scored by the Panthers. The problem is asking for the value of $y$. \begin{align} x+y &= 34 \\ x-y &= 14 \\ 2x &= 48 \\ x &= 24 \\ y &= \fbox{10} \\ \end{align}

AMC10 First Half

AMC10 B

AMC

AMC10

10A

N/A

The sum of three numbers is $20$. The first is four times the sum of the other two. The second is seven times the third. What is the product of all three?

28

Let the numbers be $x$, $y$, and $z$ in that order. The given tells us that \begin{eqnarray}y&=&7z\\ x&=&4(y+z)=4(7z+z)=4(8z)=32z\\ x+y+z&=&32z+7z+z=40z=20\\ z&=&\frac{20}{40}=\frac{1}{2}\\ y&=&7z=7\cdot\frac{1}{2}=\frac{7}{2}\\ x&=&32z=32\cdot\frac{1}{2}=16 \end{eqnarray} Therefore, the product of all three numbers is $xyz=16\cdot\frac{7}{2}\cdot\frac{1}{2}=28 \Rightarrow \fbox{28}$.

AMC10 Second Half

AMC10 A

AMC

AMC10

10B

N/A

A restaurant offers three desserts, and exactly twice as many appetizers as main courses. A dinner consists of an appetizer, a main course, and a dessert. What is the least number of main courses that a restaurant should offer so that a customer could have a different dinner each night in the year $2003$?

8

Let $m$ be the number of main courses the restaurant serves, so $2m$ is the number of appetizers. Then the number of dinner combinations is $2m\times m\times3=6m^2$. Since the customer wants to eat a different dinner in all $365$ days of $2003$, we must have \begin{align} 6m^2 &\geq 365\\ m^2 &\geq 60.83\ldots.\end{align} Also, year 2003 is not a leap year, because 2003 divided by 4 does not equal an integer. The smallest integer value that satisfies this is $\fbox{8}$.

AMC10 Second Half

AMC10 B

AMC

AMC10

10A

N/A

What is the sum of the digits of the square of $\text 111111111$?

81

Using the standard multiplication algorithm, $111,111,111^2=12,345,678,987,654,321,$ whose digit sum is $\fbox{81}$

AMC10 First Half

AMC10 A