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5 values
var_a = 10 var_b = 11 var_c = 12 list_a= [] if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_d = 2 list_b=list_a.copy() list_b.sort() var_e = list_b[-var_d] var_f = 2 list_c=list_a.copy() list_c.sort() var_g = list_c[var_f-1] var_h = var_e / var_g print(int(var_h))
1
1
[OP_LIST_SOL] 10 11 12 [OP_LIST_EOL] 2 [OP_LIST_MAX] 2 [OP_LIST_MIN] [OP_DIV]
var_a = 10 var_b = 11 var_c = 12 list_a= [] if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_d = 2 list_b=list_a.copy() list_b.sort() var_e = list_b[-var_d] var_f = 2 list_c=list_a.copy() list_c.sort() var_g = list_c[var_f-1] var_h = var_e / var_g print(int(var_h))
[OP_LIST_SOL] 10 11 12 [OP_LIST_EOL] 2 [OP_LIST_MAX] 2 [OP_LIST_MIN] [OP_DIV]
When three numbers are given: 10, 11 and 12, find the 2nd largest number divided by the 2nd smallest number.
3개의 수 10, 11, 12κ°€ μžˆμŠ΅λ‹ˆλ‹€. κ·Έ μ€‘μ—μ„œ 2번째둜 큰 μˆ˜μ™€ 2번째둜 μž‘μ€ 수둜 λ‚˜λˆˆ 값은 μ–Όλ§ˆμž…λ‹ˆκΉŒ?
Arithmetic calculation
var_a = 9 var_b = 1 var_c = var_a - var_b var_d = 9 var_e = var_c + var_d print(int(var_e))
17
17
9 1 [OP_SUB] 9 [OP_ADD]
var_a = 9 var_b = 1 var_c = var_a - var_b var_d = 9 var_e = var_c + var_d print(int(var_e))
9 1 [OP_SUB] 9 [OP_ADD]
You need to find the sum of 9 and a certain number, but mistakenly calculated the difference of them and got 1 as the answer. If the number is less than 10, find the sum of 9 and the number.
9와 μ–΄λ–€ 수의 합을 ꡬ해야 ν•˜λŠ”λ° 잘λͺ»ν•˜μ—¬ μ°¨λ₯Ό κ΅¬ν–ˆλ”λ‹ˆ 1μ΄μ—ˆμŠ΅λ‹ˆλ‹€. μ–΄λ–€ μˆ˜λŠ” 10보닀 μž‘μ€ 수일 λ•Œ, 9와 μ–΄λ–€ 수의 합을 κ΅¬ν•˜μ‹œμ˜€.
Correspondence
var_a = 20 var_b = 1 var_c = var_a + var_b var_d = 25 var_e = var_c * var_d print(int(var_e))
525
525
20 1 [OP_ADD] 25 [OP_MUL]
var_a = 20 var_b = 1 var_c = var_a + var_b var_d = 25 var_e = var_c * var_d print(int(var_e))
20 1 [OP_ADD] 25 [OP_MUL]
There are a total of 20 telephone poles at 25m (m) intervals between Seokgi's and Dongmin's houses. If there is no telephone pole in front of their houses, what is the distance in meters (m) between Seokgi's and Dongmin's houses?
석기와 λ™λ―Όμ΄μ˜ 집사이에 25λ―Έν„°(m)κ°„κ²©μ˜ μ „λ΄‡λŒ€κ°€ 총 20개 μžˆμŠ΅λ‹ˆλ‹€. 석기와 동민이 μ§‘μ•žμ—λŠ” μ „λ΄‡λŒ€κ°€ 없을 μ‹œ 석기와 λ™λ―Όμ΄μ˜ 집사이 κ±°λ¦¬λŠ” λͺ‡λ―Έν„°(m)μž…λ‹ˆκΉŒ?
Arithmetic calculation
var_a = 20 var_b = 11 var_c = 1 var_a = int(var_a) var_b = int(var_b) for i, elem in enumerate(range(var_b)): var_c = var_c * (var_a-i) for i, elem in enumerate(range(var_b)): var_c = var_c / (i+1) print(int(var_c))
167960
167960
20 11 [OP_COMB]
var_a = 20 var_b = 11 var_c = 1 var_a = int(var_a) var_b = int(var_b) for i, elem in enumerate(range(var_b)): var_c = var_c * (var_a-i) for i, elem in enumerate(range(var_b)): var_c = var_c / (i+1) print(int(var_c))
20 11 [OP_COMB]
There are 20 people in Seokjin's class. If you are trying to pick 11 people to play in a soccer tournament, what are possible cases to pick a team?
석진이넀 λ°˜μ€ 20λͺ…μž…λ‹ˆλ‹€. 좕ꡬ λŒ€νšŒμ— μΆœμ „ν•  11λͺ…을 λ½‘μœΌλ €κ³  ν•  λ•Œ, νŒ€μ„ λ½‘λŠ” κ²½μš°λŠ” λͺ¨λ‘ λͺ‡ κ°€μ§€ μž…λ‹ˆκΉŒ?
Possibility
var_a = 40 var_b = 4 var_c = var_a * var_b var_d = 40 var_e = var_c - var_d print(int(var_e))
120
120
40 4 [OP_MUL] 40 [OP_SUB]
var_a = 40 var_b = 4 var_c = var_a * var_b var_d = 40 var_e = var_c - var_d print(int(var_e))
40 4 [OP_MUL] 40 [OP_SUB]
Jisoo got 30 robots. Heeyoung has four times as many robots as Jisoo. How many robots do Heeyoung and Jisoo have in total?
μ§€μˆ˜λŠ” λ‘œλ΄‡μ„ 30개 μ–»μ—ˆλ‹€. ν¬μ˜μ΄λŠ” μ§€μˆ˜λ³΄λ‹€ λ‘œλ΄‡μ„ 4λ°° 더 많이 κ°€μ§€κ³  μžˆλ‹€. ν¬μ˜μ΄μ™€ μ§€μˆ˜λŠ” 총 λͺ‡κ°œμ˜ λ‘œλ΄‡μ„ κ°€μ§€κ³  μžˆμ„κΉŒ?
Arithmetic calculation
var_a = 5 var_b = 8 var_c = var_a + var_b var_d = 1 var_e = var_c + var_d print(int(var_e))
14
14
5 8 [OP_ADD] 1 [OP_ADD]
var_a = 5 var_b = 8 var_c = var_a + var_b var_d = 1 var_e = var_c + var_d print(int(var_e))
5 8 [OP_ADD] 1 [OP_ADD]
Given that 8 people are standing between Eunjung and Yoojung, how many students are in a line when Eunjeong is the 5th student from the front and Yoojung is the last student in the line?
은정이와 μœ μ •μ΄ μ‚¬μ΄μ—λŠ” 8λͺ…이 μ„œ μžˆμŠ΅λ‹ˆλ‹€. μ€μ •μ΄λŠ” μ•žμ—μ„œ 5번째둜 μ„œ 있고 μœ μ •μ΄λŠ” 맨 뒀에 μ„œμžˆμ„ λ•Œ 쀄을 μ„œ μžˆλŠ” 학생은 λͺ¨λ‘ λͺ‡ λͺ…μž…λ‹ˆκΉŒ?
Comparison
var_a = '민수' var_b = 'μ°½ν˜„' var_c = 'μ„±ν˜„' list_a= [] if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_d = '민수' var_e = 'μ°½ν˜„' var_f = '<' var_g = 'μ°½ν˜„' var_h = 'μ„±ν˜„' var_i = '<' list_b= [] if "/" in str(var_i): var_i = eval(str(var_i)) list_b.append(var_i) if "/" in str(var_h): var_h = eval(str(var_h)) list_b.append(var_h) if "/" in str(var_g): var_g = eval(str(var_g)) list_b.append(var_g) if "/" in str(var_f): var_f = eval(str(var_f)) list_b.append(var_f) if "/" in str(var_e): var_e = eval(str(var_e)) list_b.append(var_e) if "/" in str(var_d): var_d = eval(str(var_d)) list_b.append(var_d) list_b.reverse() global item_name_index_dict items_name_list = list_a.copy() conditions = [] condition_list = list_b.copy() temp_stack = [] for index_, cond_ in enumerate(map(str, condition_list)): if cond_ in ("<", ">", "="): operand_right = temp_stack.pop() operand_left = temp_stack.pop() if cond_ == "=": cond_ = "==" conditions.append(f"{operand_left} {cond_} {operand_right}") else: if not cond_.isdigit(): cond_ = "{" + cond_ + "}" temp_stack.append(cond_) item_name_index_dict = {} for perm in itertools.permutations(range(1, len(items_name_list) + 1)): item_name_index_dict = dict(zip(items_name_list, perm)) formatted_conditions = \ [condition.format_map(item_name_index_dict) for condition in conditions] if all(map(eval, formatted_conditions)): break list_c = list(item_name_index_dict.keys()) list_c.sort(key=item_name_index_dict.get, reverse=True) var_j = 1 var_k = list_c[var_j-1] print(var_k)
μ„±ν˜„
Seonghyun
[OP_LIST_SOL] 민수 μ°½ν˜„ μ„±ν˜„ [OP_LIST_EOL] [OP_LIST_SOL] 민수 μ°½ν˜„ < μ°½ν˜„ μ„±ν˜„ < [OP_LIST_EOL] [OP_LIST_COND_MAX_MIN] 1 [OP_LIST_GET]
var_a = 'Minsu' var_b = 'Changhyun' var_c = 'Seonghyun' list_a= [] if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_d = 'Minsu' var_e = 'Changhyun' var_f = '<' var_g = 'Changhyun' var_h = 'Seonghyun' var_i = '<' list_b= [] if "/" in str(var_i): var_i = eval(str(var_i)) list_b.append(var_i) if "/" in str(var_h): var_h = eval(str(var_h)) list_b.append(var_h) if "/" in str(var_g): var_g = eval(str(var_g)) list_b.append(var_g) if "/" in str(var_f): var_f = eval(str(var_f)) list_b.append(var_f) if "/" in str(var_e): var_e = eval(str(var_e)) list_b.append(var_e) if "/" in str(var_d): var_d = eval(str(var_d)) list_b.append(var_d) list_b.reverse() global item_name_index_dict items_name_list = list_a.copy() conditions = [] condition_list = list_b.copy() temp_stack = [] for index_, cond_ in enumerate(map(str, condition_list)): if cond_ in ("<", ">", "="): operand_right = temp_stack.pop() operand_left = temp_stack.pop() if cond_ == "=": cond_ = "==" conditions.append(f"{operand_left} {cond_} {operand_right}") else: if not cond_.isdigit(): cond_ = "{" + cond_ + "}" temp_stack.append(cond_) item_name_index_dict = {} for perm in itertools.permutations(range(1, len(items_name_list) + 1)): item_name_index_dict = dict(zip(items_name_list, perm)) formatted_conditions = \ [condition.format_map(item_name_index_dict) for condition in conditions] if all(map(eval, formatted_conditions)): break list_c = list(item_name_index_dict.keys()) list_c.sort(key=item_name_index_dict.get, reverse=True) var_j = 1 var_k = list_c[var_j-1] print(var_k)
[OP_LIST_SOL] Minsu Changhyun Seonghyun [OP_LIST_EOL] [OP_LIST_SOL] Minsu Changhyun < Changhyun Seonghyun < [OP_LIST_EOL] [OP_LIST_COND_MAX_MIN] 1 [OP_LIST_GET]
Minsu, Changhyun, and Seonghyun are standing in line at the bus stop. If Minsu is standing in front of Changhyun and Changhyun is standing in front of Seonghyun, find out who is standing at the back.
민수, μ°½ν˜„, μ„±ν˜„μ΄κ°€ λ²„μŠ€μ •λ₯˜μž₯μ—μ„œ 쀄을 μ„œ μžˆλ‹€. λ―Όμˆ˜κ°€ μ°½ν˜„μ΄ 보닀 μ•žμ— μ„œ 있고 μ°½ν˜„μΈκ°€ μ„±ν˜„μ΄ 보닀 μ•žμ— μ„œ μžˆλ‹€λ©΄ κ°€μž₯ 뒀에 μ„œ μžˆλŠ” μ‚¬λžŒμ€ λˆ„κ΅¬μΈμ§€ κ΅¬ν•˜μ‹œμ˜€.
Comparison
var_a = 5 var_b = 2 var_c = 8 var_d = 0 list_a= [] if "/" in str(var_d): var_d = eval(str(var_d)) list_a.append(var_d) if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_e = 3 list_b = [str(i) for i in list_a] list_b = list(itertools.permutations(list_b, var_e)) list_b = [''.join(num_list) for num_list in list_b] list_b = [str_num for str_num in list_b if str_num[0] != '0'] list_b = [float(i) for i in list_b] var_f = 550 list_c = [] for i in list_b: if i > var_f: list_c.append(i) var_g = 850 list_d = [] for i in list_c: if i < var_g: list_d.append(i) var_h = len(list_d) print(int(var_h))
6
6
[OP_LIST_SOL] 5 2 8 0 [OP_LIST_EOL] 3 [OP_LIST_GET_PERM] 550 [OP_LIST_MORE] 850 [OP_LIST_LESS] [OP_LIST_LEN]
var_a = 5 var_b = 2 var_c = 8 var_d = 0 list_a= [] if "/" in str(var_d): var_d = eval(str(var_d)) list_a.append(var_d) if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_e = 3 list_b = [str(i) for i in list_a] list_b = list(itertools.permutations(list_b, var_e)) list_b = [''.join(num_list) for num_list in list_b] list_b = [str_num for str_num in list_b if str_num[0] != '0'] list_b = [float(i) for i in list_b] var_f = 550 list_c = [] for i in list_b: if i > var_f: list_c.append(i) var_g = 850 list_d = [] for i in list_c: if i < var_g: list_d.append(i) var_h = len(list_d) print(int(var_h))
[OP_LIST_SOL] 5 2 8 0 [OP_LIST_EOL] 3 [OP_LIST_GET_PERM] 550 [OP_LIST_MORE] 850 [OP_LIST_LESS] [OP_LIST_LEN]
Yoo-ah left one of the digits 5, 2, 8, or 0 out and tried to form a three-digit number with the remaining digits. How many of the numbers Yoo-ah made are greater than 550 and less than 850?
μœ μ•„λŠ” 5, 2, 8, 0 쀑 ν•˜λ‚˜μ˜ 숫자λ₯Ό λΉΌκ³ , 남은 μˆ«μžλ“€λ‘œ μ„Έ 자리수λ₯Ό λ§Œλ“œλ €κ³  ν•©λ‹ˆλ‹€. μœ μ•„κ°€ λ§Œλ“œλŠ” 수 쀑 550보닀 크고 850보닀 μž‘μ€ μˆ˜λŠ” λͺ¨λ‘ λͺ‡ κ°€μ§€ μž…λ‹ˆκΉŒ?
Possibility
var_a = 9 var_b = 3 var_c = var_a - var_b var_d = 1 var_e = var_c + var_d print(int(var_e))
7
7
9 3 [OP_SUB] 1 [OP_ADD]
var_a = 9 var_b = 3 var_c = var_a - var_b var_d = 1 var_e = var_c + var_d print(int(var_e))
9 3 [OP_SUB] 1 [OP_ADD]
If you draw a diagonal from one vertex of a figure made of 9 angles, how many figures made of 3 angles are made?
9개의 각으둜 이루어진 λ„ν˜•μ˜ ν•œ κΌ­μ§“μ μ—μ„œ λŒ€κ°μ„ μ„ κΈ‹λŠ”λ‹€λ©΄, 3개의 각으둜 이루어진 λ„ν˜•μ€ λͺ‡ 개 λ§Œλ“€μ–΄μ§‘λ‹ˆκΉŒ?
Geometry
var_a = 22 var_b = 13 var_c = var_a - var_b var_d = 1 var_e = var_c + var_d print(int(var_e))
10
10
22 13 [OP_SUB] 1 [OP_ADD]
var_a = 22 var_b = 13 var_c = var_a - var_b var_d = 1 var_e = var_c + var_d print(int(var_e))
22 13 [OP_SUB] 1 [OP_ADD]
22 people of different heights stand in a line in order, tallest first. (a) stands 13th from the back. If you line up again in order of shortest to tallest, what number will you stand from the back?
각각의 ν‚€κ°€ λ‹€λ₯Έ 22λͺ…이 ν‚€κ°€ 큰 μ‚¬λžŒλΆ€ν„° μˆœμ„œλŒ€λ‘œ ν•œ μ€„λ‘œ μ„œ μžˆμŠ΅λ‹ˆλ‹€. (κ°€)λŠ” λ’€μ—μ„œ 13λ²ˆμ§Έμ— μ„œ μžˆμŠ΅λ‹ˆλ‹€. ν‚€κ°€ μž‘μ€ μ‚¬λžŒλΆ€ν„° μˆœμ„œλŒ€λ‘œ λ‹€μ‹œ ν•œ μ€„λ‘œ μ„œλ©΄ (κ°€)λŠ” λ’€μ—μ„œ λͺ‡ λ²ˆμ§Έμ— μ„œκ²Œ λ©λ‹ˆκΉŒ?
Comparison
var_a = 300 var_b = 300 var_c = 4 var_d = 100 var_e = var_c / var_d var_f = var_b * var_e var_g = 100 var_h = 10 var_i = var_g / var_h var_j = var_f * var_i var_k = var_a - var_j print(int(var_k))
180
180
300 300 4 100 [OP_DIV] [OP_MUL] 100 10 [OP_DIV] [OP_MUL] [OP_SUB]
var_a = 300 var_b = 300 var_c = 4 var_d = 100 var_e = var_c / var_d var_f = var_b * var_e var_g = 100 var_h = 10 var_i = var_g / var_h var_j = var_f * var_i var_k = var_a - var_j print(int(var_k))
300 300 4 100 [OP_DIV] [OP_MUL] 100 10 [OP_DIV] [OP_MUL] [OP_SUB]
How many grams (g) of water must be evaporated from 300 grams (g) of 4% salt water to make 10% salt water?
4%의 μ†ŒκΈˆλ¬Ό 300그램(g)μ—μ„œ λͺ‡ 그램(g)의 물을 μ¦λ°œμ‹œμΌœμ•Ό 10%의 μ†ŒκΈˆλ¬Όμ΄ λ˜λŠ”μ§€ κ΅¬ν•˜μ—¬λΌ
Arithmetic calculation
var_a = '윀기' var_b = 'μ •κ΅­' list_a= [] if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_c = 4 var_d = 6 var_e = 3 var_f = var_d * var_e list_b= [] if "/" in str(var_f): var_f = eval(str(var_f)) list_b.append(var_f) if "/" in str(var_c): var_c = eval(str(var_c)) list_b.append(var_c) list_b.reverse() var_g = 1 list_c=list_b.copy() list_c.sort() var_h = list_c[var_g-1] var_i = list_b.index(var_h)+1 var_j = list_a[var_i-1] print(var_j)
윀기
Yoongi
[OP_LIST_SOL] 윀기 μ •κ΅­ [OP_LIST_EOL] [OP_LIST_SOL] 4 6 3 [OP_MUL] [OP_LIST_EOL] 1 [OP_LIST_MIN] [OP_LIST_INDEX] [OP_LIST_POP] [OP_LIST_GET]
var_a = 'Yoongi' var_b = 'Jungkook' list_a= [] if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_c = 4 var_d = 6 var_e = 3 var_f = var_d * var_e list_b= [] if "/" in str(var_f): var_f = eval(str(var_f)) list_b.append(var_f) if "/" in str(var_c): var_c = eval(str(var_c)) list_b.append(var_c) list_b.reverse() var_g = 1 list_c=list_b.copy() list_c.sort() var_h = list_c[var_g-1] var_i = list_b.index(var_h)+1 var_j = list_a[var_i-1] print(var_j)
[OP_LIST_SOL] Yoongi Jungkook [OP_LIST_EOL] [OP_LIST_SOL] 4 6 3 [OP_MUL] [OP_LIST_EOL] 1 [OP_LIST_MIN] [OP_LIST_INDEX] [OP_LIST_POP] [OP_LIST_GET]
Yoongi collected 4 apples and Jungkook collected 6 times 3 apples. Who has fewer apples?
μœ€κΈ°λŠ” 4개, μ •κ΅­μ—λŠ” 6 κ³±ν•˜κΈ° 3만큼 사과λ₯Ό λͺ¨μ•˜μŠ΅λ‹ˆλ‹€. λˆ„κ°€ 더 사과λ₯Ό 적게 κ°€μ§€κ³  μžˆμŠ΅λ‹ˆκΉŒ?
Comparison
var_a = 'μ •κ΅­' var_b = '윀기' list_a= [] if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_c = 6 var_d = 3 var_e = var_c - var_d var_f = 4 list_b= [] if "/" in str(var_f): var_f = eval(str(var_f)) list_b.append(var_f) if "/" in str(var_e): var_e = eval(str(var_e)) list_b.append(var_e) list_b.reverse() var_g = 1 list_c=list_b.copy() list_c.sort() var_h = list_c[var_g-1] var_i = list_b.index(var_h)+1 var_j = list_a[var_i-1] print(var_j)
μ •κ΅­
Jungkook
[OP_LIST_SOL] μ •κ΅­ 윀기 [OP_LIST_EOL] [OP_LIST_SOL] 6 3 [OP_SUB] 4 [OP_LIST_EOL] 1 [OP_LIST_MIN] [OP_LIST_INDEX] [OP_LIST_POP] [OP_LIST_GET]
var_a = 'Jungkook' var_b = 'Yoongi' list_a= [] if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_c = 6 var_d = 3 var_e = var_c - var_d var_f = 4 list_b= [] if "/" in str(var_f): var_f = eval(str(var_f)) list_b.append(var_f) if "/" in str(var_e): var_e = eval(str(var_e)) list_b.append(var_e) list_b.reverse() var_g = 1 list_c=list_b.copy() list_c.sort() var_h = list_c[var_g-1] var_i = list_b.index(var_h)+1 var_j = list_a[var_i-1] print(var_j)
[OP_LIST_SOL] Jungkook Yoongi [OP_LIST_EOL] [OP_LIST_SOL] 6 3 [OP_SUB] 4 [OP_LIST_EOL] 1 [OP_LIST_MIN] [OP_LIST_INDEX] [OP_LIST_POP] [OP_LIST_GET]
Jungkook ate 3 out of 6 apples. When Yoongi has 4 apples, who has fewer apples?
μ •κ΅­μ΄λŠ” 6개의 μ‚¬κ³Όμ—μ„œ 3개λ₯Ό λ¨Ήμ—ˆμŠ΅λ‹ˆλ‹€. μœ€κΈ°κ°€ κ°€μ§„ 사과가 4개라면 λˆ„κ°€ 더 적은 사과λ₯Ό κ°€μ§€κ³  μžˆλ‚˜μš”?
Comparison
var_a = 210 var_b = 4 var_c = 3 var_d = var_b + var_c var_e = var_a / var_d print(int(var_e))
30
30
210 4 3 [OP_ADD] [OP_DIV]
var_a = 210 var_b = 4 var_c = 3 var_d = var_b + var_c var_e = var_a / var_d print(int(var_e))
210 4 3 [OP_ADD] [OP_DIV]
There are apples, bananas and persimmons in the fruit shop. The number of apples is 4 times the number of bananas, and the number of persimmons is 3 times the number of bananas. If the sum of apples and persimmons is 210, how many bananas are there?
과일 κ°€κ²Œμ— 사과, λ°”λ‚˜λ‚˜, 감이 μžˆμŠ΅λ‹ˆλ‹€. 사과 μˆ˜λŠ” λ°”λ‚˜λ‚˜ 수의 4배이고, 감 μˆ˜λŠ” λ°”λ‚˜λ‚˜ 수의 3λ°°μž…λ‹ˆλ‹€. 사과와 감 수의 합이 210개라면 λ°”λ‚˜λ‚˜λŠ” λͺ‡ κ°œμΈμ§€ ꡬ해 λ³΄μ‹œμ˜€.
Arithmetic calculation
var_a = 40 var_b = 18 var_c = var_a + var_b var_d = 23 var_e = var_c - var_d print(int(var_e))
35
35
40 18 [OP_ADD] 23 [OP_SUB]
var_a = 40 var_b = 18 var_c = var_a + var_b var_d = 23 var_e = var_c - var_d print(int(var_e))
40 18 [OP_ADD] 23 [OP_SUB]
There are 23 more soccer balls at Hoseok's school than basketballs and 18 fewer volleyballs than soccer balls. If there are 40 volleyballs, how many basketballs are there?
ν˜Έμ„μ΄λ„€ 학ꡐ에 μžˆλŠ” 좕ꡬ곡은 농ꡬ곡보닀 23개 더 많고 배ꡬ곡은 좕ꡬ곡보닀 18개 더 μ μŠ΅λ‹ˆλ‹€. 배ꡬ곡이 40개 μžˆμ„ λ•Œ, 농ꡬ곡은 λͺ‡ κ°œμž…λ‹ˆκΉŒ?
Arithmetic calculation
var_a = 4.235 var_b = 0.55 var_c = var_a / var_b var_d = 0.55 var_e = var_c / var_d print(int(var_e))
14
14
4.235 0.55 [OP_DIV] 0.55 [OP_DIV]
var_a = 4.235 var_b = 0.55 var_c = var_a / var_b var_d = 0.55 var_e = var_c / var_d print(int(var_e))
4.235 0.55 [OP_DIV] 0.55 [OP_DIV]
Multiplying a number by 0.55 in a row gives you 4.235. Find the number.
μ–΄λ–€ 수λ₯Ό 0.55둜 두 번 연속 κ³±ν•˜μ˜€λ”λ‹ˆ 4.235κ°€ λ˜μ—ˆμŠ΅λ‹ˆλ‹€. μ–΄λ–€ μˆ˜λŠ” κ΅¬ν•˜μ‹œμ˜€.
Correspondence
var_a = 6 var_b = 1 var_c = var_a + var_b var_d = 5 var_e = var_c + var_d print(int(var_e))
12
12
6 1 [OP_ADD] 5 [OP_ADD]
var_a = 6 var_b = 1 var_c = var_a + var_b var_d = 5 var_e = var_c + var_d print(int(var_e))
6 1 [OP_ADD] 5 [OP_ADD]
If Eunji stands in line for a bus and there are 6 people waiting in front and 5 people behind, how many people are there?
은지가 λ²„μŠ€ 쀄을 μ„°λŠ”λ° μ•žμ— 6λͺ… 뒀에 5λͺ… μžˆλ‹€λ©΄, ν˜„μž¬ 쀄에 μžˆλŠ” μ‚¬λžŒμ€ λͺ¨λ‘ λͺ‡ λͺ…μž…λ‹ˆκΉŒ?
Comparison
var_a = 242.7 var_b = 5 var_c = var_a * var_b print('{:.2f}'.format(round(var_c+1e-10,2)))
1213.50
1213.5
242.7 5 [OP_MUL]
var_a = 242.7 var_b = 5 var_c = var_a * var_b print('{:.2f}'.format(round(var_c+1e-10,2)))
242.7 5 [OP_MUL]
The circumference of Suyeong's school playground is 242.7 meters (m). If Suyeong ran 5 laps around the playground, how many meters (m) did she run?
μˆ˜μ˜μ΄λ„€ ν•™κ΅μ˜ μš΄λ™μž₯ λ‘˜λ ˆλŠ” 242.7 λ―Έν„°(m) μž…λ‹ˆλ‹€ μˆ˜μ˜μ΄κ°€ μš΄λ™μž₯을 5 바퀴 달렸닀면, λͺ¨λ‘ λͺ‡ λ―Έν„°(m)λ₯Ό 달린 것 μž…λ‹ˆκΉŒ?
Arithmetic calculation
var_a = 'λΉ¨κ°„' var_b = 'νŒŒλž€' var_c = '초둝' list_a= [] if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_d = 3 var_e = 2 var_f = 5 list_b= [] if "/" in str(var_f): var_f = eval(str(var_f)) list_b.append(var_f) if "/" in str(var_e): var_e = eval(str(var_e)) list_b.append(var_e) if "/" in str(var_d): var_d = eval(str(var_d)) list_b.append(var_d) list_b.reverse() var_g = 1 list_c=list_b.copy() list_c.sort() var_h = list_c[-var_g] var_i = list_b.index(var_h)+1 var_j = list_a[var_i-1] print(var_j)
초둝
green
[OP_LIST_SOL] λΉ¨κ°„ νŒŒλž€ 초둝 [OP_LIST_EOL] [OP_LIST_SOL] 3 2 5 [OP_LIST_EOL] 1 [OP_LIST_MAX] [OP_LIST_INDEX] [OP_LIST_POP] [OP_LIST_GET]
var_a = 'red' var_b = 'blue' var_c = 'green' list_a= [] if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_d = 3 var_e = 2 var_f = 5 list_b= [] if "/" in str(var_f): var_f = eval(str(var_f)) list_b.append(var_f) if "/" in str(var_e): var_e = eval(str(var_e)) list_b.append(var_e) if "/" in str(var_d): var_d = eval(str(var_d)) list_b.append(var_d) list_b.reverse() var_g = 1 list_c=list_b.copy() list_c.sort() var_h = list_c[-var_g] var_i = list_b.index(var_h)+1 var_j = list_a[var_i-1] print(var_j)
[OP_LIST_SOL] red blue green [OP_LIST_EOL] [OP_LIST_SOL] 3 2 5 [OP_LIST_EOL] 1 [OP_LIST_MAX] [OP_LIST_INDEX] [OP_LIST_POP] [OP_LIST_GET]
There are 3 red balls, 2 blue balls, and 5 green balls in the bag. When a ball is drawn from the bag, what will be the color of the ball with the highest probability?
μ£Όλ¨Έλ‹ˆ 속에 λΉ¨κ°„ 곡이 3개, νŒŒλž€ 곡이 2개, 초둝 곡이 5개 λ“€μ–΄μžˆλ‹€. μ£Όλ¨Έλ‹ˆμ—μ„œ 곡 1개λ₯Ό κΊΌλ‚Ό λ•Œ, ν™•λ₯ μ΄ κ°€μž₯ 높은 곡의 색깔을 μ“°μ‹œμ˜€.
Comparison
var_a = 3 var_b = 5 var_c = 8 var_d = 9 list_a= [] if "/" in str(var_d): var_d = eval(str(var_d)) list_a.append(var_d) if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_e = 3 list_b = [str(i) for i in list_a] list_b = list(itertools.permutations(list_b, var_e)) list_b = [''.join(num_list) for num_list in list_b] list_b = [str_num for str_num in list_b if str_num[0] != '0'] list_b = [float(i) for i in list_b] var_f = len(list_b) print(int(var_f))
24
24
[OP_LIST_SOL] 3 5 8 9 [OP_LIST_EOL] 3 [OP_LIST_GET_PERM] [OP_LIST_LEN]
var_a = 3 var_b = 5 var_c = 8 var_d = 9 list_a= [] if "/" in str(var_d): var_d = eval(str(var_d)) list_a.append(var_d) if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_e = 3 list_b = [str(i) for i in list_a] list_b = list(itertools.permutations(list_b, var_e)) list_b = [''.join(num_list) for num_list in list_b] list_b = [str_num for str_num in list_b if str_num[0] != '0'] list_b = [float(i) for i in list_b] var_f = len(list_b) print(int(var_f))
[OP_LIST_SOL] 3 5 8 9 [OP_LIST_EOL] 3 [OP_LIST_GET_PERM] [OP_LIST_LEN]
Soyun chooses and discards one of the number cards 3, 5, 8, and 9, and wants to make a three-digit natural number with the remaining cards. How many total numbers can Soyun make?
μ†Œμœ€μ΄λŠ” 숫자 μΉ΄λ“œ 3, 5, 8, 9쀑 ν•˜λ‚˜λ₯Ό 골라 버리고, 남은 μΉ΄λ“œλ“€λ‘œ μ„Έ 자리 μžμ—°μˆ˜λ₯Ό λ§Œλ“€κ³ μž ν•©λ‹ˆλ‹€. μ†Œμœ€μ΄κ°€ λ§Œλ“€ 수 μžˆλŠ” μˆ˜λ“€μ˜ κ°œμˆ˜λŠ” 총 λͺ‡κ°œμž…λ‹ˆκΉŒ?
Possibility
var_a = 0 var_b = 3 var_c = 5 var_d = 9 list_a= [] if "/" in str(var_d): var_d = eval(str(var_d)) list_a.append(var_d) if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_e = 3 list_b = [str(i) for i in list_a] list_b = list(itertools.permutations(list_b, var_e)) list_b = [''.join(num_list) for num_list in list_b] list_b = [str_num for str_num in list_b if str_num[0] != '0'] list_b = [float(i) for i in list_b] var_f = 1 list_c=list_b.copy() list_c.sort() var_g = list_c[var_f-1] var_h = 1 list_d=list_b.copy() list_d.sort() var_i = list_d[-var_h] var_j = var_g + var_i print(int(var_j))
1258
1258
[OP_LIST_SOL] 0 3 5 9 [OP_LIST_EOL] 3 [OP_LIST_GET_PERM] 1 [OP_LIST_MIN] 1 [OP_LIST_MAX] [OP_ADD]
var_a = 0 var_b = 3 var_c = 5 var_d = 9 list_a= [] if "/" in str(var_d): var_d = eval(str(var_d)) list_a.append(var_d) if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_e = 3 list_b = [str(i) for i in list_a] list_b = list(itertools.permutations(list_b, var_e)) list_b = [''.join(num_list) for num_list in list_b] list_b = [str_num for str_num in list_b if str_num[0] != '0'] list_b = [float(i) for i in list_b] var_f = 1 list_c=list_b.copy() list_c.sort() var_g = list_c[var_f-1] var_h = 1 list_d=list_b.copy() list_d.sort() var_i = list_d[-var_h] var_j = var_g + var_i print(int(var_j))
[OP_LIST_SOL] 0 3 5 9 [OP_LIST_EOL] 3 [OP_LIST_GET_PERM] 1 [OP_LIST_MIN] 1 [OP_LIST_MAX] [OP_ADD]
If you choose three different numbers from 0, 3, 5, and 9 and use them once to make a three-digit number, what is the sum of the smallest and largest numbers?
0, 3, 5, 9 쀑 각각 λ‹€λ₯Έ 3개λ₯Ό 골라 ν•œ λ²ˆμ”© μ‚¬μš©ν•˜μ—¬ μ„Έ 자리 수λ₯Ό λ§Œλ“ λ‹€κ³  ν•  λ•Œ, κ°€μž₯ μž‘μ€ μˆ˜μ™€ κ°€μž₯ 큰 수의 합은 μ–Όλ§ˆμΈκ°€?
Possibility
var_a = 1.625 var_b = 1000 var_c = var_a * var_b var_d = 115 var_e = var_c - var_d var_f = 400 var_g = var_e / var_f print('{:.2f}'.format(round(var_g+1e-10,2)))
3.78
3.78
1.625 1000 [OP_MUL] 115 [OP_SUB] 400 [OP_DIV]
var_a = 1.625 var_b = 1000 var_c = var_a * var_b var_d = 115 var_e = var_c - var_d var_f = 400 var_g = var_e / var_f print('{:.2f}'.format(round(var_g+1e-10,2)))
1.625 1000 [OP_MUL] 115 [OP_SUB] 400 [OP_DIV]
There is a hot air balloon that is 115 meters (m) high. How many minutes does it take to pass through the stratosphere of 1.625 kilometers (km) if it ascends 400 meters (m) in 1 minute?
높이가 115λ―Έν„°(m)인 열기ꡬ가 μžˆμŠ΅λ‹ˆλ‹€. 1뢄에 400λ―Έν„°(m)λ₯Ό μ˜¬λΌκ°„λ‹€κ³  ν•  λ•Œ 1.625ν‚¬λ‘œλ―Έν„°(㎞)의 μ„±μΈ΅κΆŒμ„ ν†΅κ³Όν•˜λŠ”λ° λͺ‡ 뢄이 κ±Έλ¦¬λ‚˜μš”?
Arithmetic calculation
var_a = '영주' var_b = '민호' list_a= [] if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_c = 97.5 var_d = 97 var_e = 3 var_f = 10 var_g = var_e / var_f var_h = var_d + var_g list_b= [] if "/" in str(var_h): var_h = eval(str(var_h)) list_b.append(var_h) if "/" in str(var_c): var_c = eval(str(var_c)) list_b.append(var_c) list_b.reverse() var_i = 1 list_c=list_b.copy() list_c.sort() var_j = list_c[var_i-1] var_k = list_b.index(var_j)+1 var_l = list_a[var_k-1] print(var_l)
민호
Minho
[OP_LIST_SOL] 영주 민호 [OP_LIST_EOL] [OP_LIST_SOL] 97.5 97 3 10 [OP_DIV] [OP_ADD] [OP_LIST_EOL] 1 [OP_LIST_MIN] [OP_LIST_INDEX] [OP_LIST_POP] [OP_LIST_GET]
var_a = 'Yeongju' var_b = 'Minho' list_a= [] if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_c = 97.5 var_d = 97 var_e = 3 var_f = 10 var_g = var_e / var_f var_h = var_d + var_g list_b= [] if "/" in str(var_h): var_h = eval(str(var_h)) list_b.append(var_h) if "/" in str(var_c): var_c = eval(str(var_c)) list_b.append(var_c) list_b.reverse() var_i = 1 list_c=list_b.copy() list_c.sort() var_j = list_c[var_i-1] var_k = list_b.index(var_j)+1 var_l = list_a[var_k-1] print(var_l)
[OP_LIST_SOL] Yeongju Minho [OP_LIST_EOL] [OP_LIST_SOL] 97.5 97 3 10 [OP_DIV] [OP_ADD] [OP_LIST_EOL] 1 [OP_LIST_MIN] [OP_LIST_INDEX] [OP_LIST_POP] [OP_LIST_GET]
The wire Yeongju has is 97.5 centimeters (cm), and the wire Minho has is 3 millimeters (mm) longer than 97 centimeters (cm). Who has a shorter wire?
μ˜μ£Όκ°€ κ°€μ§€κ³  μžˆλŠ” μ² μ‚¬λŠ” 97.5μ„Όν‹°λ―Έν„°(㎝)이고, λ―Όν˜Έκ°€ κ°€μ§€κ³  μžˆλŠ” μ² μ‚¬λŠ” 97μ„Όν‹°λ―Έν„°(㎝)보닀 3밀리미터(㎜)더 κΉλ‹ˆλ‹€. 더 짧은 철사λ₯Ό κ°€μ§€κ³  μžˆλŠ” μ‚¬λžŒμ€ λˆ„κ΅¬μž…λ‹ˆκΉŒ?
Comparison
var_a = 1620006 var_b = 1629996 var_c = 1 list_a = [i for i in range(var_a, var_b + 1, var_c)] var_d = 123 list_b = [] var_d = int(var_d) for i in list_a: i = int(i) if i % var_d == 0: list_b.append(i) var_e = 10 var_f = 6 list_c = [] var_e = int(var_e) var_f = int(var_f) if var_f < 0: var_f = var_f + var_e for i in list_b: i = int(i) if i%var_e == var_f: list_c.append(i) var_g = 1 list_d=list_c.copy() list_d.sort() var_h = list_d[var_g-1] print(int(var_h))
1620156
1620156
1620006 1629996 1 [OP_LIST_ARANGE] 123 [OP_LIST_DIVISIBLE] 10 6 [OP_LIST_DIVIDE_AND_REMAIN] 1 [OP_LIST_MIN]
var_a = 1620006 var_b = 1629996 var_c = 1 list_a = [i for i in range(var_a, var_b + 1, var_c)] var_d = 123 list_b = [] var_d = int(var_d) for i in list_a: i = int(i) if i % var_d == 0: list_b.append(i) var_e = 10 var_f = 6 list_c = [] var_e = int(var_e) var_f = int(var_f) if var_f < 0: var_f = var_f + var_e for i in list_b: i = int(i) if i%var_e == var_f: list_c.append(i) var_g = 1 list_d=list_c.copy() list_d.sort() var_h = list_d[var_g-1] print(int(var_h))
1620006 1629996 1 [OP_LIST_ARANGE] 123 [OP_LIST_DIVISIBLE] 10 6 [OP_LIST_DIVIDE_AND_REMAIN] 1 [OP_LIST_MIN]
Among the multiples of 123, there are is a number shaped162ABC6. Find the smallest of these numbers shaped like that.
123의 배수 쀑 162ABC6 λͺ¨μ–‘μ˜ μˆ˜κ°€ μžˆλ‹€. μ΄λŸ¬ν•œ 수 μ€‘μ—μ„œ κ°€μž₯ μž‘μ€ 수λ₯Ό κ΅¬ν•˜μ—¬λΌ.
Correspondence
var_a = 2 var_b = 11 var_c = var_a * var_b var_d = 6 var_e = var_c * var_d print(int(var_e))
132
132
2 11 [OP_MUL] 6 [OP_MUL]
var_a = 2 var_b = 11 var_c = var_a * var_b var_d = 6 var_e = var_c * var_d print(int(var_e))
2 11 [OP_MUL] 6 [OP_MUL]
Divide 11 by a number and you get 2 as a result. How much is 6 multiplied by a number?
11을 μ–΄λ–€ μˆ˜μ—μ„œ λ‚˜λˆ„μ—ˆλ”λ‹ˆ 결과둜 2λ₯Ό μ–»μ—ˆμŠ΅λ‹ˆλ‹€. 6을 μ–΄λ–€ μˆ˜μ— κ³±ν•˜λ©΄ μ–Όλ§ˆμž…λ‹ˆκΉŒ?
Correspondence
var_a = 3.8 var_b = 2 var_c = 45 var_d = 60 var_e = var_c / var_d var_f = var_b + var_e var_g = var_a * var_f print('{:.2f}'.format(round(var_g+1e-10,2)))
10.45
10.45
3.8 2 45 60 [OP_DIV] [OP_ADD] [OP_MUL]
var_a = 3.8 var_b = 2 var_c = 45 var_d = 60 var_e = var_c / var_d var_f = var_b + var_e var_g = var_a * var_f print('{:.2f}'.format(round(var_g+1e-10,2)))
3.8 2 45 60 [OP_DIV] [OP_ADD] [OP_MUL]
Hot air balloons are said to ascend into the sky at 3.8 kilometers (km) per hour. How high in kilometers (km) does a hot air balloon rise without stopping in 2 hours and 45 minutes?
μ—΄κΈ°κ΅¬λŠ” ν•œ μ‹œκ°„μ— 3.8ν‚¬λ‘œλ―Έν„°(㎞)λ₯Ό ν•˜λŠ˜λ‘œ μ˜¬λΌκ°„λ‹€κ³  ν•©λ‹ˆλ‹€. 열기ꡬ가 2μ‹œκ°„ 45λΆ„λ™μ•ˆ μ‰¬μ§€μ•Šκ³  μ˜¬λΌκ°„ λ†’μ΄λŠ” λͺ‡ ν‚¬λ‘œλ―Έν„°(㎞)μž…λ‹ˆκΉŒ?
Arithmetic calculation
var_a = 12 var_b = 400 var_c = 1000 var_d = var_b / var_c var_e = var_a + var_d var_f = 2 var_g = var_e / var_f var_h = 2 var_i = 600 var_j = 1000 var_k = var_i / var_j var_l = var_h + var_k var_m = 2 var_n = var_l / var_m var_o = var_g + var_n print('{:.2f}'.format(round(var_o+1e-10,2)))
7.5
7.5
12 400 1000 [OP_DIV] [OP_ADD] 2 [OP_DIV] 2 600 1000 [OP_DIV] [OP_ADD] 2 [OP_DIV] [OP_ADD]
var_a = 12 var_b = 400 var_c = 1000 var_d = var_b / var_c var_e = var_a + var_d var_f = 2 var_g = var_e / var_f var_h = 2 var_i = 600 var_j = 1000 var_k = var_i / var_j var_l = var_h + var_k var_m = 2 var_n = var_l / var_m var_o = var_g + var_n print('{:.2f}'.format(round(var_o+1e-10,2)))
12 400 1000 [OP_DIV] [OP_ADD] 2 [OP_DIV] 2 600 1000 [OP_DIV] [OP_ADD] 2 [OP_DIV] [OP_ADD]
There are 12 liters (L) and 400 milliliters (γŽ–) of oil. You are trying to divide this into two bottles. A large bottle can hold 2 liters (L) and 600 milliliters (γŽ–) more than a small bottle. How many liters (liters) will the large bottle hold?
12리터(L) 400밀리리터(γŽ–)의 기름이 μžˆμŠ΅λ‹ˆλ‹€. 이것을 큰 λ³‘μ—λŠ” μž‘μ€ 병보닀 2리터(L) 600밀리리터(γŽ–) 많게 λ‚˜λˆ„μ–΄ λ‹΄μœΌλ €κ³  ν•©λ‹ˆλ‹€. 큰 λ³‘μ—λŠ” λͺ‡ 리터(L)λ₯Ό λ‹΄κ²Œ λ©λ‹ˆκΉŒ?
Arithmetic calculation
var_a = 10 var_b = 3 var_c = var_a - var_b print(int(var_c))
7
7
10 3 [OP_SUB]
var_a = 10 var_b = 3 var_c = var_a - var_b print(int(var_c))
10 3 [OP_SUB]
How many diagonals can be drawn from one vertex of a 10-angle figure?
10개의 각으둜 이루어진 λ„ν˜•μ˜ ν•œ κΌ­μ§“μ μ—μ„œ 그을 수 μžˆλŠ” λŒ€κ°μ„ μ˜ κ°œμˆ˜λŠ” λͺ‡ κ°œμž…λ‹ˆκΉŒ?
Geometry
var_a = 'A375B' ans_dict = dict() var_a = str(var_a) list_a = [] variable_candi = set(['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z']) for v in set(var_a): if v in variable_candi: ans_dict[v] = 0 candi = list(itertools.product('0123456789', repeat=len(ans_dict))) for c in candi: temp = var_a for i, (k, _) in enumerate(ans_dict.items()): temp = temp.replace(k, str(c[i])) if len(var_a) == len(str(int(temp))): new_elem = int(temp) list_a.append(new_elem) var_b = 24 list_b = [] var_b = int(var_b) for i in list_a: i = int(i) if i % var_b == 0: list_b.append(i) var_c = 'A375B' var_d = 'A' var_c = str(var_c) var_d = str(var_d) unk_idx = var_c.index(var_d) list_c = [] for elem in list_b: elem = str(elem) list_c.append(int(elem[unk_idx])) list_c = list(set(list_c)) var_e = len(list_c) print(int(var_e))
3
3
A375B [OP_GEN_POSSIBLE_LIST] 24 [OP_LIST_DIVISIBLE] A375B A [OP_LIST_FIND_UNK] [OP_LIST_LEN]
var_a = 'A375B' ans_dict = dict() var_a = str(var_a) list_a = [] variable_candi = set(['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z']) for v in set(var_a): if v in variable_candi: ans_dict[v] = 0 candi = list(itertools.product('0123456789', repeat=len(ans_dict))) for c in candi: temp = var_a for i, (k, _) in enumerate(ans_dict.items()): temp = temp.replace(k, str(c[i])) if len(var_a) == len(str(int(temp))): new_elem = int(temp) list_a.append(new_elem) var_b = 24 list_b = [] var_b = int(var_b) for i in list_a: i = int(i) if i % var_b == 0: list_b.append(i) var_c = 'A375B' var_d = 'A' var_c = str(var_c) var_d = str(var_d) unk_idx = var_c.index(var_d) list_c = [] for elem in list_b: elem = str(elem) list_c.append(int(elem[unk_idx])) list_c = list(set(list_c)) var_e = len(list_c) print(int(var_e))
A375B [OP_GEN_POSSIBLE_LIST] 24 [OP_LIST_DIVISIBLE] A375B A [OP_LIST_FIND_UNK] [OP_LIST_LEN]
The five-digit number A375B is divisible by 24. How many numbers fit for A?
λ‹€μ„― 자리 수 A375BλŠ” 24둜 λ‚˜λˆ„μ–΄ λ–¨μ–΄μ§„λ‹€. A에 μ•Œλ§žμ€ μˆ«μžλŠ” λͺ¨λ‘ λͺ‡ κ°œμΈκ°€?
Correspondence
var_a = 21 var_b = 3 var_c = var_a / var_b var_d = 2 var_e = var_c * var_d print(int(var_e))
14
14
21 3 [OP_DIV] 2 [OP_MUL]
var_a = 21 var_b = 3 var_c = var_a / var_b var_d = 2 var_e = var_c * var_d print(int(var_e))
21 3 [OP_DIV] 2 [OP_MUL]
What is the total number of vertices of a figure with 21 edges, 9 faces, and two polygonal base sides that are parallel and congruent?
λͺ¨μ„œλ¦¬κ°€ 21개, 면이 9개이고, λ‹€κ°ν˜•μΈ 두 밑면이 μ„œλ‘œ ν‰ν–‰ν•˜κ³  합동인 λ„ν˜•μ˜ κΌ­μ§“μ μ˜ κ°œμˆ˜λŠ” 총 λͺ‡ κ°œμž…λ‹ˆκΉŒ?
Geometry
var_a = 0.8 var_b = 0.5 var_c = 0.9 var_d = 0.3333333333333333 list_a= [] if "/" in str(var_d): var_d = eval(str(var_d)) list_a.append(var_d) if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_e = 3 list_b = [] for i in list_a: if i <= var_e: list_b.append(i) var_f = len(list_b) print(int(var_f))
4
4
[OP_LIST_SOL] 0.8 1/2 0.9 1/3 [OP_LIST_EOL] 3 [OP_LIST_LESS_EQUAL] [OP_LIST_LEN]
var_a = 0.8 var_b = 0.5 var_c = 0.9 var_d = 0.3333333333333333 list_a= [] if "/" in str(var_d): var_d = eval(str(var_d)) list_a.append(var_d) if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_e = 3 list_b = [] for i in list_a: if i <= var_e: list_b.append(i) var_f = len(list_b) print(int(var_f))
[OP_LIST_SOL] 0.8 1/2 0.9 1/3 [OP_LIST_EOL] 3 [OP_LIST_LESS_EQUAL] [OP_LIST_LEN]
There are four numbers 0.8, 1/2, 0.9, and 1/3. How many numbers are less than or equal to 3?
4개의 숫자 0.8, 1/2, 0.9, 1/3κ°€ μžˆμŠ΅λ‹ˆλ‹€. 3보닀 μž‘κ±°λ‚˜ 같은 μˆ˜λŠ” λͺ‡ κ°œμž…λ‹ˆκΉŒ?
Comparison
var_a = 'A15B94' ans_dict = dict() var_a = str(var_a) list_a = [] variable_candi = set(['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z']) for v in set(var_a): if v in variable_candi: ans_dict[v] = 0 candi = list(itertools.product('0123456789', repeat=len(ans_dict))) for c in candi: temp = var_a for i, (k, _) in enumerate(ans_dict.items()): temp = temp.replace(k, str(c[i])) if len(var_a) == len(str(int(temp))): new_elem = int(temp) list_a.append(new_elem) var_b = 99 list_b = [] var_b = int(var_b) for i in list_a: i = int(i) if i % var_b == 0: list_b.append(i) var_c = 'A15B94' var_d = 'A' var_c = str(var_c) var_d = str(var_d) unk_idx = var_c.index(var_d) var_e = 0 for elem in list_b: elem = str(elem) var_e = int(elem[unk_idx]) var_f = 'A15B94' var_g = 'B' var_f = str(var_f) var_g = str(var_g) unk_idx = var_f.index(var_g) var_h = 0 for elem in list_b: elem = str(elem) var_h = int(elem[unk_idx]) var_i = var_e + var_h print(int(var_i))
8
8
A15B94 [OP_GEN_POSSIBLE_LIST] 99 [OP_LIST_DIVISIBLE] A15B94 A [OP_LIST_FIND_UNK] A15B94 B [OP_LIST_FIND_UNK] [OP_ADD]
var_a = 'A15B94' ans_dict = dict() var_a = str(var_a) list_a = [] variable_candi = set(['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z']) for v in set(var_a): if v in variable_candi: ans_dict[v] = 0 candi = list(itertools.product('0123456789', repeat=len(ans_dict))) for c in candi: temp = var_a for i, (k, _) in enumerate(ans_dict.items()): temp = temp.replace(k, str(c[i])) if len(var_a) == len(str(int(temp))): new_elem = int(temp) list_a.append(new_elem) var_b = 99 list_b = [] var_b = int(var_b) for i in list_a: i = int(i) if i % var_b == 0: list_b.append(i) var_c = 'A15B94' var_d = 'A' var_c = str(var_c) var_d = str(var_d) unk_idx = var_c.index(var_d) var_e = 0 for elem in list_b: elem = str(elem) var_e = int(elem[unk_idx]) var_f = 'A15B94' var_g = 'B' var_f = str(var_f) var_g = str(var_g) unk_idx = var_f.index(var_g) var_h = 0 for elem in list_b: elem = str(elem) var_h = int(elem[unk_idx]) var_i = var_e + var_h print(int(var_i))
A15B94 [OP_GEN_POSSIBLE_LIST] 99 [OP_LIST_DIVISIBLE] A15B94 A [OP_LIST_FIND_UNK] A15B94 B [OP_LIST_FIND_UNK] [OP_ADD]
If the six-digit number A15B94 is a multiple of 99, find the sum of A and B.
μ—¬μ„― 자리 수 A15B94κ°€ 99의 배수일 λ•Œ, A와 B의 합을 κ΅¬ν•˜μ‹œμ˜€.
Correspondence
var_a = 24 var_b = 34 var_c = 4 var_d = var_b // var_c var_e = var_a / var_d var_f = 4 var_g = var_e * var_f var_h = 34 var_i = var_g + var_h var_j = 2 var_k = var_i * var_j print(int(var_k))
92
92
24 34 4 [OP_FDIV] [OP_DIV] 4 [OP_MUL] 34 [OP_ADD] 2 [OP_MUL]
var_a = 24 var_b = 34 var_c = 4 var_d = var_b // var_c var_e = var_a / var_d var_f = 4 var_g = var_e * var_f var_h = 34 var_i = var_g + var_h var_j = 2 var_k = var_i * var_j print(int(var_k))
24 34 4 [OP_FDIV] [OP_DIV] 4 [OP_MUL] 34 [OP_ADD] 2 [OP_MUL]
You have a rectangular piece of dough measuring 34 centimeters (cm) wide. When the dough is cut without gaps using a square mold with a side length of 4 centimeters (cm), if the width of the dough is 2 centimeters (cm) left and 24 cookies are made in total, Find the minimum perimeter of the dough before the mold is stamped in centimeters (cm).
κ°€λ‘œμ˜ 길이가 34μ„Όν‹°λ―Έν„°(㎝)인 μ§μ‚¬κ°ν˜• λͺ¨μ–‘μ˜ 반죽이 μžˆμŠ΅λ‹ˆλ‹€. ν•œ λ³€μ˜ 길이가 4μ„Όν‹°λ―Έν„°(㎝)인 μ •μ‚¬κ°ν˜• λͺ¨μ–‘μœΌλ‘œ 생긴 틀을 μ΄μš©ν•΄ λ°˜μ£½μ„ λΉˆν‹ˆμ—†μ΄ μ°μ—ˆμ„ λ•Œ, κ°€λ‘œμ˜ 길이가 2μ„Όν‹°λ―Έν„°(㎝) 남고, μΏ ν‚€κ°€ 총 24개 λ§Œλ“€μ–΄μ‘Œλ‹€λ©΄, 틀을 찍기 μ „μ˜ 반죽의 μ΅œμ†Œ λ‘˜λ ˆλŠ” λͺ‡ μ„Όν‹°λ―Έν„°(㎝)인지 κ΅¬ν•˜μ‹œμ˜€.
Geometry
var_a = '사과' var_b = 'λ³΅μˆ­μ•„' var_c = 'λ°°' var_d = 'λ”ΈκΈ°' list_a= [] if "/" in str(var_d): var_d = eval(str(var_d)) list_a.append(var_d) if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_e = len(list_a) var_f = 2 var_g = var_e + var_f var_h = 1 var_i = var_g - var_h var_j = 2 var_k = 1 var_i = int(var_i) var_j = int(var_j) for i, elem in enumerate(range(var_j)): var_k = var_k * (var_i-i) for i, elem in enumerate(range(var_j)): var_k = var_k / (i+1) print(int(var_k))
10
10
[OP_LIST_SOL] 사과 λ³΅μˆ­μ•„ λ°° λ”ΈκΈ° [OP_LIST_EOL] [OP_LIST_LEN] 2 [OP_ADD] 1 [OP_SUB] 2 [OP_COMB]
var_a = 'apples' var_b = 'peaches' var_c = 'pears' var_d = 'strawberries' list_a= [] if "/" in str(var_d): var_d = eval(str(var_d)) list_a.append(var_d) if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_e = len(list_a) var_f = 2 var_g = var_e + var_f var_h = 1 var_i = var_g - var_h var_j = 2 var_k = 1 var_i = int(var_i) var_j = int(var_j) for i, elem in enumerate(range(var_j)): var_k = var_k * (var_i-i) for i, elem in enumerate(range(var_j)): var_k = var_k / (i+1) print(int(var_k))
[OP_LIST_SOL] apples peaches pears strawberries [OP_LIST_EOL] [OP_LIST_LEN] 2 [OP_ADD] 1 [OP_SUB] 2 [OP_COMB]
At a fruit store that sells apples, peaches, pears, and strawberries, you want to buy two fruits allowing duplicates. How many possible cases are there?
사과, λ³΅μˆ­μ•„, λ°°, λ”ΈκΈ°λ₯Ό νŒλ§€ν•˜λŠ” κ³ΌμΌκ°€κ²Œμ—μ„œ 쀑볡을 ν—ˆλ½ν•˜μ—¬ 2개의 과일을 κ΅¬λ§€ν•˜λ €κ³  ν•©λ‹ˆλ‹€. κ°€λŠ₯ν•œ 경우의 μˆ˜λŠ” λͺ‡ κ°€μ§€ μΌκΉŒμš”?
Possibility
var_a = 200 var_b = 200 var_c = 5 var_d = 100 var_e = var_c / var_d var_f = var_b * var_e var_g = 100 var_h = 8 var_i = var_g / var_h var_j = var_f * var_i var_k = var_a - var_j print(int(var_k))
75
75
200 200 5 100 [OP_DIV] [OP_MUL] 100 8 [OP_DIV] [OP_MUL] [OP_SUB]
var_a = 200 var_b = 200 var_c = 5 var_d = 100 var_e = var_c / var_d var_f = var_b * var_e var_g = 100 var_h = 8 var_i = var_g / var_h var_j = var_f * var_i var_k = var_a - var_j print(int(var_k))
200 200 5 100 [OP_DIV] [OP_MUL] 100 8 [OP_DIV] [OP_MUL] [OP_SUB]
You have 200 grams (g) of salt water with 5% salt. When the remaining salt became 8% of the total salt water after the water evaporated from this salt water, find how many grams (g) of water has evaporated.
μ†ŒκΈˆμ΄ 5% λ“€μ–΄μžˆλŠ” μ†ŒκΈˆλ¬Όμ΄ 200그램(g) μžˆμŠ΅λ‹ˆλ‹€. 이 μ†ŒκΈˆλ¬Όμ—μ„œ 물이 μ¦λ°œν•œ ν›„ 남은 μ†ŒκΈˆμ€ 전체 μ†ŒκΈˆλ¬Όμ˜ 8%κ°€ λ˜μ—ˆμ„ λ•Œ, 물은 λͺ‡ 그램(g) μ¦λ°œν•˜μ˜€λŠ”μ§€ κ΅¬ν•˜μ‹œμ˜€.
Arithmetic calculation
var_a = 20 var_b = 2 var_c = var_a + var_b var_d = 6 var_e = var_c * var_d var_f = 28 var_g = var_e * var_f print(int(var_g))
3696
3696
20 2 [OP_ADD] 6 [OP_MUL] 28 [OP_MUL]
var_a = 20 var_b = 2 var_c = var_a + var_b var_d = 6 var_e = var_c * var_d var_f = 28 var_g = var_e * var_f print(int(var_g))
20 2 [OP_ADD] 6 [OP_MUL] 28 [OP_MUL]
There are 28 identical bookshelves in the library. Each bookshelf has 6 floors, and the number of books on each floor is the same. It is said that when two books are taken from one floor of a bookshelf, there are 20 books left on that floor. How many books are on the shelves in the library?
λ„μ„œκ΄€μ— λ˜‘κ°™μ€ μ±…μž₯이 28개 μžˆμŠ΅λ‹ˆλ‹€. 각 μ±…μž₯은 6측이고, 각 μΈ΅λ§ˆλ‹€ κ½‚ν˜€ μžˆλŠ” μ±…μ˜ μˆ˜λŠ” κ°™μŠ΅λ‹ˆλ‹€. μ–΄λŠ μ±…μž₯의 ν•œ μΈ΅μ—μ„œ 2개의 책을 λ½‘μ•˜μ„ λ•Œ κ·Έ 측에 20ꢌ이 λ‚¨μ•˜λ‹€κ³  ν•©λ‹ˆλ‹€. λ„μ„œκ΄€μ˜ μ±…μž₯에 κ½‚ν˜€ μžˆλŠ” 책은 λͺ¨λ‘ λͺ‡ κΆŒμž…λ‹ˆκΉŒ?
Comparison
var_a = 15 var_b = 182 var_c = var_a * var_b var_d = 105 var_e = var_c // var_d print(int(var_e))
26
26
15 182 [OP_MUL] 105 [OP_FDIV]
var_a = 15 var_b = 182 var_c = var_a * var_b var_d = 105 var_e = var_c // var_d print(int(var_e))
15 182 [OP_MUL] 105 [OP_FDIV]
In the process of division, Seokjin left out the 0 in the tens place of the three-digit subtracting number, so he saw it as 15 by mistake. If he gets a quotient of 182 without any remainder, find the quotient of the correct calculation.
석진이가 λ‚˜λˆ—μ…ˆμ„ ν•˜λŠ” κ³Όμ •μ—μ„œ λ‚˜λˆ„λŠ” μ„Έ 자리 수의 μ‹­μ˜ μžλ¦¬μ— μžˆλŠ” 0을 λΉ λœ¨λ €μ„œ 15둜 잘λͺ» 보고 κ³„μ‚°ν•˜μ˜€λ”λ‹ˆ λͺ«μ΄ 182둜 λ‚˜λˆ„μ–΄λ–¨μ–΄μ‘ŒμŠ΅λ‹ˆλ‹€. λ°”λ₯΄κ²Œ κ³„μ‚°ν•œ λͺ«μ„ κ΅¬ν•˜μ‹œμ˜€.
Arithmetic calculation
var_a = 1.4 var_b = 0.9 var_c = 1.2 var_d = 0.5 var_e = 1.3 list_a= [] if "/" in str(var_e): var_e = eval(str(var_e)) list_a.append(var_e) if "/" in str(var_d): var_d = eval(str(var_d)) list_a.append(var_d) if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_f = 1.1 list_b = [] for i in list_a: if i >= var_f: list_b.append(i) var_g = 1 list_c=list_b.copy() list_c.sort() var_h = list_c[var_g-1] print('{:.2f}'.format(round(var_h+1e-10,2)))
1.20
1.2
[OP_LIST_SOL] 1.4 9/10 1.2 0.5 13/10 [OP_LIST_EOL] 1.1 [OP_LIST_MORE_EQUAL] 1 [OP_LIST_MIN]
var_a = 1.4 var_b = 0.9 var_c = 1.2 var_d = 0.5 var_e = 1.3 list_a= [] if "/" in str(var_e): var_e = eval(str(var_e)) list_a.append(var_e) if "/" in str(var_d): var_d = eval(str(var_d)) list_a.append(var_d) if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_f = 1.1 list_b = [] for i in list_a: if i >= var_f: list_b.append(i) var_g = 1 list_c=list_b.copy() list_c.sort() var_h = list_c[var_g-1] print('{:.2f}'.format(round(var_h+1e-10,2)))
[OP_LIST_SOL] 1.4 9/10 1.2 0.5 13/10 [OP_LIST_EOL] 1.1 [OP_LIST_MORE_EQUAL] 1 [OP_LIST_MIN]
Which of the five numbers 1.4, 9/10, 1.2, 0.5, and 13/10 is greater than or equal to 1.1? Find the smallest or the equal number.
5κ°€μ§€ 숫자 1.4, 9/10, 1.2, 0.5, 13/10 μ€‘μ—μ„œ 1.1보닀 ν¬κ±°λ‚˜ 같은 수 쀑 κ°€μž₯ μž‘κ±°λ‚˜ 같은 μˆ«μžλŠ” λ¬΄μ—‡μž…λ‹ˆκΉŒ?
Comparison
var_a = 55 var_b = 17 var_c = var_a + var_b var_d = 9 var_e = var_c / var_d print(int(var_e))
8
8
55 17 [OP_ADD] 9 [OP_DIV]
var_a = 55 var_b = 17 var_c = var_a + var_b var_d = 9 var_e = var_c / var_d print(int(var_e))
55 17 [OP_ADD] 9 [OP_DIV]
Subtracting 17 from a number gives 55. What number is it when the number is divided by 9?
μ–΄λ–€ μˆ˜μ— 17을 λΊλ”λ‹ˆ κ²°κ³Όκ°€ 55κ°€ λ˜μ—ˆμŠ΅λ‹ˆλ‹€. 9λ₯Ό μ–΄λ–€ 수둜 λ‚˜λˆ„λ©΄ μ–Όλ§ˆμž…λ‹ˆκΉŒ?
Correspondence
var_a = 3 print(int(var_a))
3
3
3
var_a = 3 print(int(var_a))
3
How many faces in a tetrahedron meet at one vertex?
μ •μ‚¬λ©΄μ²΄μ—μ„œ ν•œ 꼭짓점에 λͺ¨μΈ 면의 μˆ˜λŠ” λͺ¨λ‘ λͺ‡κ°œμΈκ°€μš”?
Possibility
var_a = 28 var_b = 2 var_c = var_a / var_b var_d = 6 var_e = var_c - var_d var_f = 6 var_g = var_e * var_f print(int(var_g))
48
48
28 2 [OP_DIV] 6 [OP_SUB] 6 [OP_MUL]
var_a = 28 var_b = 2 var_c = var_a / var_b var_d = 6 var_e = var_c - var_d var_f = 6 var_g = var_e * var_f print(int(var_g))
28 2 [OP_DIV] 6 [OP_SUB] 6 [OP_MUL]
Using 28 centimeters (cm) long wire, I made a rectangle with a width of 6. Find the area of the rectangle.
길이가 28μ„Όν‹°λ―Έν„°(㎝)인 철사λ₯Ό μ‚¬μš©ν•˜μ—¬ κ°€λ‘œμ˜ 길이가 6인 μ§μ‚¬κ°ν˜•μ„ λ§Œλ“€μ—ˆμŠ΅λ‹ˆλ‹€. μ§μ‚¬κ°ν˜•μ˜ 넓이λ₯Ό κ΅¬ν•˜μ‹œμ˜€.
Geometry
var_a = 819 var_b = 57 var_c = var_a + var_b var_d = 6 var_e = 9 var_f = var_d - var_e var_g = var_c - var_f var_h = 57 var_i = var_g - var_h print(int(var_i))
822
822
819 57 [OP_ADD] 6 9 [OP_SUB] [OP_SUB] 57 [OP_SUB]
var_a = 819 var_b = 57 var_c = var_a + var_b var_d = 6 var_e = 9 var_f = var_d - var_e var_g = var_c - var_f var_h = 57 var_i = var_g - var_h print(int(var_i))
819 57 [OP_ADD] 6 9 [OP_SUB] [OP_SUB] 57 [OP_SUB]
While Yoongi was subtracting 57 from the three-digit number, Yoongi mistakenly saw the three-digit ones digit 9 as 6. If the difference that Yoongi found is 819, what is the correct calculation?
μœ€κΈ°λŠ” μ„Έ 자리 μˆ˜μ— 57을 빼던 쀑, μœ€κΈ°λŠ” μ„Έ 자리 수의 일의 자리 숫자 9λ₯Ό 6으둜 잘λͺ» λ΄€μŠ΅λ‹ˆλ‹€. μœ€κΈ°κ°€ κ΅¬ν•œ μ°¨κ°€ 819라고 ν•˜λ©΄, λ°”λ₯΄κ²Œ κ³„μ‚°ν•œ κ²°κ³ΌλŠ” λ¬΄μ—‡μΈκ°€μš”?
Correspondence
var_a = 3 var_b = 2 var_c = 1 var_a = int(var_a) var_b = int(var_b) for i, elem in enumerate(range(var_b)): var_c = var_c * (var_a-i) print(int(var_c))
6
6
3 2 [OP_PERM]
var_a = 3 var_b = 2 var_c = 1 var_a = int(var_a) var_b = int(var_b) for i, elem in enumerate(range(var_b)): var_c = var_c * (var_a-i) print(int(var_c))
3 2 [OP_PERM]
Find the number of semi-straight lines for three points that are not on a straight line.
ν•œ 직선 μœ„μ— μžˆμ§€ μ•Šμ€ μ„Έ 점에 λŒ€ν•˜μ—¬ λ°˜μ§μ„ μ˜ 개수λ₯Ό κ΅¬ν•˜μ‹œμ˜€.
Geometry
var_a = 17 var_b = 17 var_c = 5 var_d = var_b * var_c var_e = 11 var_f = var_d - var_e var_g = var_a + var_f var_h = 17 var_i = 5 var_j = var_h * var_i var_k = 11 var_l = var_j - var_k var_m = 2 var_n = var_l * var_m var_o = 8 var_p = var_n + var_o var_q = var_g + var_p print(int(var_q))
247
247
17 17 5 [OP_MUL] 11 [OP_SUB] [OP_ADD] 17 5 [OP_MUL] 11 [OP_SUB] 2 [OP_MUL] 8 [OP_ADD] [OP_ADD]
var_a = 17 var_b = 17 var_c = 5 var_d = var_b * var_c var_e = 11 var_f = var_d - var_e var_g = var_a + var_f var_h = 17 var_i = 5 var_j = var_h * var_i var_k = 11 var_l = var_j - var_k var_m = 2 var_n = var_l * var_m var_o = 8 var_p = var_n + var_o var_q = var_g + var_p print(int(var_q))
17 17 5 [OP_MUL] 11 [OP_SUB] [OP_ADD] 17 5 [OP_MUL] 11 [OP_SUB] 2 [OP_MUL] 8 [OP_ADD] [OP_ADD]
When three people have marbles, 5 times the number of marbles that Namjoon has is 11 more than what Yoongi has. 2 multiply by the number that Yoongi has is 8 fewer than Hoseok's marbles. Find the sum of marbles for each person: Namjoon, Yoongi, and Hoseok.
남쀀이가 κ°€μ§„ ꡬ슬 수의 5λ°°λŠ” μœ€κΈ°κ°€ κ°€μ§„ ꡬ슬 개수 보닀 11개 더 λ§ŽμŠ΅λ‹ˆλ‹€. μœ€κΈ°κ°€ κ°€μ§„ ꡬ슬 수의 2λ°°λŠ” ν˜Έμ„μ΄κ°€ κ°€μ§„ ꡬ슬 개수 보닀 8개 더 μ μŠ΅λ‹ˆλ‹€. 남쀀, 윀기, ν˜Έμ„μ΄κ°€ κ°€μ§€κ³  μžˆλŠ” κ΅¬μŠ¬μ€ λͺ¨λ‘ λͺ‡ κ°œμž…λ‹ˆκΉŒ?
Arithmetic calculation
var_a = 900 var_b = 7 var_c = var_a * var_b var_d = 200 var_e = var_c + var_d var_f = 1000 var_g = var_e / var_f print('{:.2f}'.format(round(var_g+1e-10,2)))
6.50
6.5
900 7 [OP_MUL] 200 [OP_ADD] 1000 [OP_DIV]
var_a = 900 var_b = 7 var_c = var_a * var_b var_d = 200 var_e = var_c + var_d var_f = 1000 var_g = var_e / var_f print('{:.2f}'.format(round(var_g+1e-10,2)))
900 7 [OP_MUL] 200 [OP_ADD] 1000 [OP_DIV]
A large bucket is filled with water. You want to divide this water into 7 smaller buckets of the same shape and size. If you divide it by 900 grams (g), 200 grams (g) remain. How many kilograms (kg) of water is in the large bucket?
큰 톡에 물이 가득 λ“€μ–΄ μžˆμŠ΅λ‹ˆλ‹€. 이 물을 λͺ¨μ–‘κ³Ό 크기가 같은 7개의 μž‘μ€ 물톡에 λ‚˜λˆ„μ–΄ λ‹΄μœΌλ €κ³  ν•©λ‹ˆλ‹€. 900그램(g)μ”© λ‚˜λˆ„μ–΄ λ‹΄μœΌλ©΄ 200그램(g)이 λ‚¨μŠ΅λ‹ˆλ‹€. 큰 물톡에 λ“€μ–΄ μžˆλŠ” 물은 λͺ‡ ν‚¬λ‘œκ·Έλž¨(㎏)μž…λ‹ˆκΉŒ?
Arithmetic calculation
var_a = 1 var_b = 125 var_c = 1 list_a = [i for i in range(var_a, var_b + 1, var_c)] var_d="" for i in list_a: i = str(i) var_d = var_d + i list_b = [] var_d = int(var_d) while var_d//10 > 0: list_b.append(var_d%10) var_d = var_d//10 list_b.append(var_d%10) list_b = list_b[::-1] var_e = 2 var_f = 0 var_e = int(var_e) for i in list_b: i = int(i) if i == var_e: var_f = var_f + 1 print(int(var_f))
29
29
1 125 1 [OP_LIST_ARANGE] [OP_LIST2NUM] [OP_NUM2LIST] 2 [OP_LIST_FIND_NUM]
var_a = 1 var_b = 125 var_c = 1 list_a = [i for i in range(var_a, var_b + 1, var_c)] var_d="" for i in list_a: i = str(i) var_d = var_d + i list_b = [] var_d = int(var_d) while var_d//10 > 0: list_b.append(var_d%10) var_d = var_d//10 list_b.append(var_d%10) list_b = list_b[::-1] var_e = 2 var_f = 0 var_e = int(var_e) for i in list_b: i = int(i) if i == var_e: var_f = var_f + 1 print(int(var_f))
1 125 1 [OP_LIST_ARANGE] [OP_LIST2NUM] [OP_NUM2LIST] 2 [OP_LIST_FIND_NUM]
When writing numbers from 1 to 125, find out how many times 2 are written.
1λΆ€ν„° 125κΉŒμ§€ 수λ₯Ό μ“Έ λ•Œ, 0λΆ€ν„° 9κΉŒμ§€μ˜ 숫자 쀑 2λŠ” λͺ¨λ‘ λͺ‡ 번 μ“°λŠ”μ§€ κ΅¬ν•˜μ‹œμ˜€.
Arithmetic calculation
var_a = 16 var_b = 7 var_c = var_a * var_b var_d = 13 var_e = var_c - var_d var_f = 8 var_g = var_e % var_f print(int(var_g))
3
3
16 7 [OP_MUL] 13 [OP_SUB] 8 [OP_MOD]
var_a = 16 var_b = 7 var_c = var_a * var_b var_d = 13 var_e = var_c - var_d var_f = 8 var_g = var_e % var_f print(int(var_g))
16 7 [OP_MUL] 13 [OP_SUB] 8 [OP_MOD]
I bought 7 boxes of soda, each box containing 16 cans. Thirteen of them are discarded because the cans were crushed during transport, and I plan to distribute the remaining beverages equally among the eight cartons. How many cans of beverages do not fit in the box?
ν•œ μƒμžμ— 16μΊ”μ”© λ“€μ–΄μžˆλŠ” 음료수λ₯Ό 7μƒμž μƒ€μŠ΅λ‹ˆλ‹€. κ·Έμ€‘μ—μ„œ 13κ°œλŠ” 운반 도쀑 캔이 μ°Œκ·ΈλŸ¬μ Έμ„œ 버리고, 남은 음료수λ₯Ό 8μƒμžμ— λ˜‘κ°™μ΄ λ‚˜λˆ„μ–΄ λ‹΄μœΌλ €κ³  ν•©λ‹ˆλ‹€. μƒμžμ— λ“€μ–΄κ°€μ§€ λͺ»ν•˜λŠ” μŒλ£Œμˆ˜λŠ” λͺ‡ μΊ”μž…λ‹ˆκΉŒ?
Arithmetic calculation
var_a = 5 var_b = 7 var_c = var_a + var_b var_d = 2 var_e = var_c // var_d print(int(var_e))
6
6
5 7 [OP_ADD] 2 [OP_FDIV]
var_a = 5 var_b = 7 var_c = var_a + var_b var_d = 2 var_e = var_c // var_d print(int(var_e))
5 7 [OP_ADD] 2 [OP_FDIV]
In a running competition, Jungkook came in the 7th and Minyoung came in the 5th. Taehyung did worse than Minyoung but better than Jungkook. In what place did Taehyung come in?
달리기 μ‹œν•©μ—μ„œ μ •κ΅­μ΄λŠ” 7등을 ν–ˆκ³ , λ―Όμ˜μ΄λŠ” 5등을 ν–ˆμŠ΅λ‹ˆλ‹€. νƒœν˜•μ΄λŠ” λ―Όμ˜μ΄λ³΄λ‹€ λͺ»ν–ˆμ§€λ§Œ μ •κ΅­μ΄λ³΄λ‹€λŠ” μž˜ν–ˆμŠ΅λ‹ˆλ‹€. νƒœν˜•μ΄λŠ” λͺ‡ λ“±μž…λ‹ˆκΉŒ?
Comparison
var_a = 19 var_b = 24 var_c = var_a * var_b print(int(var_c))
456
456
19 24 [OP_MUL]
var_a = 19 var_b = 24 var_c = var_a * var_b print(int(var_c))
19 24 [OP_MUL]
When a number is divided by 24, it becomes 19. What number is it?
μ–΄λ–€ 수λ₯Ό 24둜 λ‚˜λˆ„μ—ˆλ”λ‹ˆ λͺ«μ΄ 19 μ˜€μŠ΅λ‹ˆλ‹€. μ–΄λ–€ μˆ˜λŠ” μ–Όλ§ˆμΌκΉŒμš”?
Correspondence
var_a = 8 var_b = 1 var_c = var_a + var_b var_d = 8 var_e = 2 var_f = var_d * var_e var_g = var_c + var_f var_h = 8 var_i = 1 var_j = var_h + var_i var_k = var_g + var_j print(int(var_k))
34
34
8 1 [OP_ADD] 8 2 [OP_MUL] [OP_ADD] 8 1 [OP_ADD] [OP_ADD]
var_a = 8 var_b = 1 var_c = var_a + var_b var_d = 8 var_e = 2 var_f = var_d * var_e var_g = var_c + var_f var_h = 8 var_i = 1 var_j = var_h + var_i var_k = var_g + var_j print(int(var_k))
8 1 [OP_ADD] 8 2 [OP_MUL] [OP_ADD] 8 1 [OP_ADD] [OP_ADD]
Find the sum of the vertices, faces, and edges of the octagonal pyramid.
νŒ”κ°λΏ”μ˜ 꼭짓점, λ©΄, λͺ¨μ„œλ¦¬μ˜ 수의 합을 κ΅¬ν•˜μ„Έμš”.
Geometry
var_a = 8 var_b = 9 var_c = var_a * var_b var_d = 3 var_e = var_c - var_d var_f = 8 var_g = 9 var_h = var_f * var_g var_i = 3 var_j = var_h - var_i var_k = 3 var_l = var_j / var_k var_m = 8 var_n = 9 var_o = var_m * var_n var_p = 3 var_q = var_o - var_p var_r = 3 var_s = var_q / var_r var_t = 2 var_u = var_s * var_t var_v = 4 var_w = var_u + var_v list_a= [] if "/" in str(var_w): var_w = eval(str(var_w)) list_a.append(var_w) if "/" in str(var_l): var_l = eval(str(var_l)) list_a.append(var_l) if "/" in str(var_e): var_e = eval(str(var_e)) list_a.append(var_e) list_a.reverse() var_x = 1 list_b=list_a.copy() list_b.sort() var_y = list_b[-var_x] var_z = 3 list_c=list_a.copy() list_c.sort() var_A = list_c[-var_z] var_B = var_y + var_A print(int(var_B))
92
92
[OP_LIST_SOL] 8 9 [OP_MUL] 3 [OP_SUB] 8 9 [OP_MUL] 3 [OP_SUB] 3 [OP_DIV] 8 9 [OP_MUL] 3 [OP_SUB] 3 [OP_DIV] 2 [OP_MUL] 4 [OP_ADD] [OP_LIST_EOL] 1 [OP_LIST_MAX] 3 [OP_LIST_MAX] [OP_ADD]
var_a = 8 var_b = 9 var_c = var_a * var_b var_d = 3 var_e = var_c - var_d var_f = 8 var_g = 9 var_h = var_f * var_g var_i = 3 var_j = var_h - var_i var_k = 3 var_l = var_j / var_k var_m = 8 var_n = 9 var_o = var_m * var_n var_p = 3 var_q = var_o - var_p var_r = 3 var_s = var_q / var_r var_t = 2 var_u = var_s * var_t var_v = 4 var_w = var_u + var_v list_a= [] if "/" in str(var_w): var_w = eval(str(var_w)) list_a.append(var_w) if "/" in str(var_l): var_l = eval(str(var_l)) list_a.append(var_l) if "/" in str(var_e): var_e = eval(str(var_e)) list_a.append(var_e) list_a.reverse() var_x = 1 list_b=list_a.copy() list_b.sort() var_y = list_b[-var_x] var_z = 3 list_c=list_a.copy() list_c.sort() var_A = list_c[-var_z] var_B = var_y + var_A print(int(var_B))
[OP_LIST_SOL] 8 9 [OP_MUL] 3 [OP_SUB] 8 9 [OP_MUL] 3 [OP_SUB] 3 [OP_DIV] 8 9 [OP_MUL] 3 [OP_SUB] 3 [OP_DIV] 2 [OP_MUL] 4 [OP_ADD] [OP_LIST_EOL] 1 [OP_LIST_MAX] 3 [OP_LIST_MAX] [OP_ADD]
There are three numbers A, B, and C. A is 8 multiplied by 9 minus 3. B is the quotient of A divided by 3. C is equal to 2 times B plus 4. Find the sum of the largest and smallest numbers.
μ„Έ 수 A, B, Cκ°€ μžˆμŠ΅λ‹ˆλ‹€. AλŠ” 8에 9λ₯Ό κ³±ν•œ ν›„ 3을 λΊ€ μˆ˜μž…λ‹ˆλ‹€. BλŠ” Aλ₯Ό 3으둜 λ‚˜λˆˆ λͺ«μž…λ‹ˆλ‹€. CλŠ” Bλ₯Ό 2λ°° ν•œ λ‹€μŒ 4λ₯Ό λ”ν•œ μˆ˜μž…λ‹ˆλ‹€. κ°€μž₯ 큰 μˆ˜μ™€ κ°€μž₯ μž‘μ€ 수의 합을 κ΅¬ν•˜μ‹œμ˜€.
Comparison
var_a = 3 var_b = 5 var_c = 7 var_d = 9 list_a= [] if "/" in str(var_d): var_d = eval(str(var_d)) list_a.append(var_d) if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_e = 4 list_b = [str(i) for i in list_a] list_b = list(itertools.permutations(list_b, var_e)) list_b = [''.join(num_list) for num_list in list_b] list_b = [str_num for str_num in list_b if str_num[0] != '0'] list_b = [float(i) for i in list_b] list_b = [float(i) for i in list_b] var_f = sum(list_b) print(int(var_f))
159984
159984
[OP_LIST_SOL] 3 5 7 9 [OP_LIST_EOL] 4 [OP_LIST_GET_PERM] [OP_LIST_SUM]
var_a = 3 var_b = 5 var_c = 7 var_d = 9 list_a= [] if "/" in str(var_d): var_d = eval(str(var_d)) list_a.append(var_d) if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_e = 4 list_b = [str(i) for i in list_a] list_b = list(itertools.permutations(list_b, var_e)) list_b = [''.join(num_list) for num_list in list_b] list_b = [str_num for str_num in list_b if str_num[0] != '0'] list_b = [float(i) for i in list_b] list_b = [float(i) for i in list_b] var_f = sum(list_b) print(int(var_f))
[OP_LIST_SOL] 3 5 7 9 [OP_LIST_EOL] 4 [OP_LIST_GET_PERM] [OP_LIST_SUM]
Find the sum of possible four digit numbers that can be formed by using each 3, 5, 7, and 9.
λ„€ 개의 숫자 3, 5, 7, 9λ₯Ό 각각 ν•œ λ²ˆμ”©λ§Œ μ‚¬μš©ν•˜μ—¬ λ§Œλ“€ 수 μžˆλŠ” λ„€ 자리 수의 총합을 κ΅¬ν•˜μ—¬λΌ.
Possibility
var_a = 5 var_b = 1 var_c = var_a - var_b var_d = 2 var_e = 1 var_f = var_d - var_e var_g = 1 var_h = var_f + var_g var_i = 1 var_c = int(var_c) var_h = int(var_h) for i, elem in enumerate(range(var_h)): var_i = var_i * (var_c-i) print(int(var_i))
12
12
5 1 [OP_SUB] 2 1 [OP_SUB] 1 [OP_ADD] [OP_PERM]
var_a = 5 var_b = 1 var_c = var_a - var_b var_d = 2 var_e = 1 var_f = var_d - var_e var_g = 1 var_h = var_f + var_g var_i = 1 var_c = int(var_c) var_h = int(var_h) for i, elem in enumerate(range(var_h)): var_i = var_i * (var_c-i) print(int(var_i))
5 1 [OP_SUB] 2 1 [OP_SUB] 1 [OP_ADD] [OP_PERM]
You are going to clean the classroom with 5 friends including Eunhee. When picking 2 people in charge of the classroom floor and 1 person in charge of the window, find the number of cases where Eunhee be in charge of the classroom floor.
은희λ₯Ό ν¬ν•¨ν•˜μ—¬ 5λͺ…μ˜ μΉœκ΅¬λ“€κ³Ό ν•¨κ»˜ ꡐ싀 μ²­μ†Œλ₯Ό ν•˜λ € ν•œλ‹€. ꡐ싀 λ°”λ‹₯ λ‹΄λ‹Ή 2λͺ…, μ°½λ¬Έ λ‹΄λ‹Ή 1λͺ…을 λ½‘λŠ”λ‹€κ³  ν•  λ•Œ, 은희가 ꡐ싀 λ°”λ‹₯ 담당이 λ˜λŠ” 경우의 수λ₯Ό κ΅¬ν•˜μ‹œμ˜€.
Possibility
var_a = 94.2 var_b = 30 var_c = var_a / var_b print('{:.2f}'.format(round(var_c+1e-10,2)))
3.14
3.14
94.2 30 [OP_DIV]
var_a = 94.2 var_b = 30 var_c = var_a / var_b print('{:.2f}'.format(round(var_c+1e-10,2)))
94.2 30 [OP_DIV]
A lake has a circumference of 30 meters (m) and a circumference of 94.2 meters (m). How many times the circumference of the lake is greater than its diameter?
μ–΄λŠ 호수의 λ‘˜λ ˆλŠ” 30λ―Έν„°(m)이며, λ‘˜λ ˆλŠ” 94.2λ―Έν„°(m)이닀. 호수의 λ‘˜λ ˆλŠ” μ§€λ¦„μ˜ λͺ‡ 배인가?
Geometry
var_a = 10 var_b = 11 var_c = 12 list_a= [] if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_d = 3 list_b=list_a.copy() list_b.sort() var_e = list_b[var_d-1] var_f = 2 list_c=list_a.copy() list_c.sort() var_g = list_c[var_f-1] var_h = var_e - var_g print(int(var_h))
1
1
[OP_LIST_SOL] 10 11 12 [OP_LIST_EOL] 3 [OP_LIST_MIN] 2 [OP_LIST_MIN] [OP_SUB]
var_a = 10 var_b = 11 var_c = 12 list_a= [] if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_d = 3 list_b=list_a.copy() list_b.sort() var_e = list_b[var_d-1] var_f = 2 list_c=list_a.copy() list_c.sort() var_g = list_c[var_f-1] var_h = var_e - var_g print(int(var_h))
[OP_LIST_SOL] 10 11 12 [OP_LIST_EOL] 3 [OP_LIST_MIN] 2 [OP_LIST_MIN] [OP_SUB]
There are three numbers 10, 11 and 12. What is the value of difference between the 3rd smallest number and the 2nd smallest number?
3개의 수 10, 11, 12κ°€ μžˆμŠ΅λ‹ˆλ‹€. κ·Έ μ€‘μ—μ„œ 3번째둜 μž‘μ€ μˆ˜μ™€ 2번째둜 μž‘μ€ 수의 μ°¨λŠ” μ–Όλ§ˆμž…λ‹ˆκΉŒ?
Arithmetic calculation
var_a = 30 var_b = 6 var_c = var_a * var_b var_d = 7 var_e = 6 var_f = 1 var_g = var_e - var_f var_h = var_d * var_g var_i = var_c - var_h var_j = 30 var_k = var_i * var_j print(int(var_k))
4350
4350
30 6 [OP_MUL] 7 6 1 [OP_SUB] [OP_MUL] [OP_SUB] 30 [OP_MUL]
var_a = 30 var_b = 6 var_c = var_a * var_b var_d = 7 var_e = 6 var_f = 1 var_g = var_e - var_f var_h = var_d * var_g var_i = var_c - var_h var_j = 30 var_k = var_i * var_j print(int(var_k))
30 6 [OP_MUL] 7 6 1 [OP_SUB] [OP_MUL] [OP_SUB] 30 [OP_MUL]
Sumin has 6 square-shaped pieces of paper with each side measuring 30 centimeters (cm). To connect these papers in a single line, she overlapped them by 7 centimeters (cm) and attached them with glue. Find the area of the connected paper.
μˆ˜λ―Όμ΄λŠ” ν•œ λ³€μ˜ 길이가 30μ„Όν‹°λ―Έν„°(㎝)인 μ •μ‚¬κ°ν˜• λͺ¨μ–‘μ˜ 쒅이λ₯Ό 6μž₯ κ°€μ§€κ³  μžˆμŠ΅λ‹ˆλ‹€. 이 쒅이λ₯Ό ν•œ μ€„λ‘œ μ—°κ²°ν•˜κΈ° μœ„ν•΄ 7μ„Όν‹°λ―Έν„°(㎝)μ”© 겹쳐 놓고 ν’€λ‘œ λΆ™μ˜€μŠ΅λ‹ˆλ‹€. μ—°κ²°λœ μ’…μ΄μ˜ 넓이λ₯Ό κ΅¬ν•˜μ‹œμ˜€.
Geometry
var_a = 7 var_b = 2 var_c = 1 var_d = var_b * var_c var_e = var_a - var_d var_f = 1 var_g = var_e + var_f print(int(var_g))
6
6
7 2 1 [OP_MUL] [OP_SUB] 1 [OP_ADD]
var_a = 7 var_b = 2 var_c = 1 var_d = var_b * var_c var_e = var_a - var_d var_f = 1 var_g = var_e + var_f print(int(var_g))
7 2 1 [OP_MUL] [OP_SUB] 1 [OP_ADD]
You want to divide 7 bananas into 2 different baskets. The basket must contain at least 1 banana. How many ways are there to share bananas?
λ°”λ‚˜λ‚˜ 7개λ₯Ό μ„œλ‘œ λ‹€λ₯Έ 2개의 λ°”κ΅¬λ‹ˆμ— λ‚˜λˆ„μ–΄ λ‹΄μœΌλ €κ³  ν•©λ‹ˆλ‹€. λ°”κ΅¬λ‹ˆμ—λŠ” 적어도 λ°”λ‚˜λ‚˜ 1개λ₯Ό λ‹΄μ•„μ•Ό ν•©λ‹ˆλ‹€. λ°”λ‚˜λ‚˜λ₯Ό λ‚˜λˆ„μ–΄ λ‹΄λŠ” 방법은 λͺ¨λ‘ λͺ‡ κ°€μ§€μž…λ‹ˆκΉŒ?
Possibility
var_a = 120 var_b = 36 var_c = var_a - var_b var_d = 2 var_e = var_c / var_d var_f = 36 var_g = var_e + var_f print(int(var_g))
78
78
120 36 [OP_SUB] 2 [OP_DIV] 36 [OP_ADD]
var_a = 120 var_b = 36 var_c = var_a - var_b var_d = 2 var_e = var_c / var_d var_f = 36 var_g = var_e + var_f print(int(var_g))
120 36 [OP_SUB] 2 [OP_DIV] 36 [OP_ADD]
Some school divides classes into upper and lower classes according to math skills. If there are a total of 120 students and there are 36 more students in the lower class than in the upper class, how many students are in the lower class?
μ–΄λŠ ν•™κ΅λŠ” μˆ˜ν•™ μ‹€λ ₯에 따라 μƒλ°˜κ³Ό ν•˜λ°˜μœΌλ‘œ λ‚˜λˆ„μ–΄μ„œ μˆ˜μ—…μ„ ν•œλ‹€. 학생이 총 120λͺ…이고 ν•˜λ°˜ 학생이 μƒλ°˜ 학생보닀 36λͺ… 더 λ§Žλ‹€λ©΄ ν•˜λ°˜ 학생은 λͺ‡ λͺ…μž…λ‹ˆκΉŒ?
Arithmetic calculation
var_a = 2 var_b = 4 var_c = var_a + var_b var_d = 1 var_e = var_c + var_d var_f = 8 var_g = var_e * var_f print(int(var_g))
56
56
2 4 [OP_ADD] 1 [OP_ADD] 8 [OP_MUL]
var_a = 2 var_b = 4 var_c = var_a + var_b var_d = 1 var_e = var_c + var_d var_f = 8 var_g = var_e * var_f print(int(var_g))
2 4 [OP_ADD] 1 [OP_ADD] 8 [OP_MUL]
There are 2 people standing in front of Jungkook and four people behind him. If there are a total of 8 lines with the same number of people for each line, how many people are there?
μ •κ΅­ μ•žμ— 2λͺ…이 μ„œ 있고, μ •κ΅­ 뒀에 4λͺ…이 μ„œμžˆλ‹€. 총 8쀄이 있고 각 쀄에 μ„œ μžˆλŠ” μ‚¬λžŒμ˜ μˆ˜κ°€ κ°™λ‹€λ©΄ 총 λͺ‡ λͺ…μž…λ‹ˆκΉŒ?
Comparison
var_a = 0.12 var_b = 0.2 var_c = var_a + var_b var_d = 55.21 var_e = var_c + var_d print('{:.2f}'.format(round(var_e+1e-10,2)))
55.53
55.53
3/25 1/5 [OP_ADD] 55.21 [OP_ADD]
var_a = 0.12 var_b = 0.2 var_c = var_a + var_b var_d = 55.21 var_e = var_c + var_d print('{:.2f}'.format(round(var_e+1e-10,2)))
3/25 1/5 [OP_ADD] 55.21 [OP_ADD]
There are 3 numbers: 3/25, 1/5, and 55.21. What is the sum of these numbers?
3/25, 1/5, 55.21 총 3개의 μˆ«μžκ°€ μžˆμŠ΅λ‹ˆλ‹€. 숫자의 합은 μ–Όλ§ˆμΈκ°€μš”?
Arithmetic calculation
var_a = 5 var_b = 5 var_c = var_a + var_b print(int(var_c))
10
10
5 5 [OP_ADD]
var_a = 5 var_b = 5 var_c = var_a + var_b print(int(var_c))
5 5 [OP_ADD]
If A and B are the numbers of sides and vertices of a pentagon, what is the value of A+B?
μ˜€κ°ν˜•μ˜ λ³€κ³Ό κΌ­μ§“μ μ˜ 개수λ₯Ό 각각 A, B라 ν•  λ•Œ, A+B의 값은?
Geometry
var_a = 1 var_b = 9 var_c = 1 list_a = [i for i in range(var_a, var_b + 1, var_c)] var_d = 3 list_b = [str(i) for i in list_a] list_b = list(itertools.permutations(list_b, var_d)) list_b = [''.join(num_list) for num_list in list_b] list_b = [str_num for str_num in list_b if str_num[0] != '0'] list_b = [float(i) for i in list_b] list_c=[] for i in list_b: var_e = 0 i = int(i) while i//10 > 0: var_e = var_e + i%10 i = i//10 var_e = var_e + i%10 list_c.append(var_e) var_f = 19 var_g = 0 var_f = int(var_f) for i in list_c: i = int(i) if i == var_f: var_g = var_g + 1 print(int(var_g))
30
30
1 9 1 [OP_LIST_ARANGE] 3 [OP_LIST_GET_PERM] [OP_LIST_NUM2SUM] 19 [OP_LIST_FIND_NUM]
var_a = 1 var_b = 9 var_c = 1 list_a = [i for i in range(var_a, var_b + 1, var_c)] var_d = 3 list_b = [str(i) for i in list_a] list_b = list(itertools.permutations(list_b, var_d)) list_b = [''.join(num_list) for num_list in list_b] list_b = [str_num for str_num in list_b if str_num[0] != '0'] list_b = [float(i) for i in list_b] list_c=[] for i in list_b: var_e = 0 i = int(i) while i//10 > 0: var_e = var_e + i%10 i = i//10 var_e = var_e + i%10 list_c.append(var_e) var_f = 19 var_g = 0 var_f = int(var_f) for i in list_c: i = int(i) if i == var_f: var_g = var_g + 1 print(int(var_g))
1 9 1 [OP_LIST_ARANGE] 3 [OP_LIST_GET_PERM] [OP_LIST_NUM2SUM] 19 [OP_LIST_FIND_NUM]
Find the number of combinations of numbers that add up to 19 by selecting 3 of the number cards from 1 to 9.
1λΆ€ν„° 9κΉŒμ§€μ˜ μˆ˜κ°€ 적힌 숫자 μΉ΄λ“œλ“€ 쀑 3μž₯을 κ³¨λΌμ„œ 합이 19κ°€ 되게 ν•˜λŠ” 수의 μ‘°ν•©μ˜ 개수λ₯Ό κ΅¬ν•˜μ‹œμ˜€.
Possibility
var_a = 2.23 var_b = 3.12 var_c = 9.434 var_d = 2.453 list_a= [] if "/" in str(var_d): var_d = eval(str(var_d)) list_a.append(var_d) if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_e = 2 list_b=list_a.copy() list_b.sort() var_f = list_b[var_e-1] print('{:.2f}'.format(round(var_f+1e-10,2)))
2.45
2.45
[OP_LIST_SOL] 2.23 3.12 9.434 2.453 [OP_LIST_EOL] 2 [OP_LIST_MIN]
var_a = 2.23 var_b = 3.12 var_c = 9.434 var_d = 2.453 list_a= [] if "/" in str(var_d): var_d = eval(str(var_d)) list_a.append(var_d) if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_e = 2 list_b=list_a.copy() list_b.sort() var_f = list_b[var_e-1] print('{:.2f}'.format(round(var_f+1e-10,2)))
[OP_LIST_SOL] 2.23 3.12 9.434 2.453 [OP_LIST_EOL] 2 [OP_LIST_MIN]
Which of 2.23, 3.12, 9.434, and 2.453 is the second smallest number?
2.23, 3.12, 9.434 및 2.453쀑 2번째둜 μž‘μ€ μˆ˜λŠ”?
Comparison
var_a = 25 var_b = 9 var_c = var_a - var_b var_d = 16 var_e = var_c + var_d print(int(var_e))
32
32
25 9 [OP_SUB] 16 [OP_ADD]
var_a = 25 var_b = 9 var_c = var_a - var_b var_d = 16 var_e = var_c + var_d print(int(var_e))
25 9 [OP_SUB] 16 [OP_ADD]
Of the students who played on the playground, 16 went home, and 9 more came, resulting in a total of 25. How many students were playing on the playground at first?
μš΄λ™μž₯μ—μ„œ λ†€λ˜ 학생 μ€‘μ—μ„œ 16λͺ…이 집에 κ°€κ³ , 9λͺ…이 더 μ™€μ„œ 25λͺ…이 λ˜μ—ˆμŠ΅λ‹ˆλ‹€. μ²˜μŒμ— μš΄λ™μž₯μ—μ„œ λ†€λ˜ 학생은 λͺ‡ λͺ…μž…λ‹ˆκΉŒ?
Arithmetic calculation
var_a = 4 var_b = 6 var_c = var_a + var_b var_d = 1 var_e = var_c - var_d print(int(var_e))
9
9
4 6 [OP_ADD] 1 [OP_SUB]
var_a = 4 var_b = 6 var_c = var_a + var_b var_d = 1 var_e = var_c - var_d print(int(var_e))
4 6 [OP_ADD] 1 [OP_SUB]
Yuna and her friends are of different heights. The teacher lined up in order from the smallest height, and Yuna was 4th from the front and 6th from the back. How many Yuna and her friends are there?
μœ λ‚˜μ™€ μΉœκ΅¬λ“€μ€ μ„œλ‘œ ν‚€κ°€ λ‹€λ¦…λ‹ˆλ‹€. μ„ μƒλ‹˜κ»˜μ„œ μ•žμ—μ„œλΆ€ν„° ν‚€κ°€ μž‘μ€ μˆœμ„œλŒ€λ‘œ 쀄을 μ„Έμ› λ”λ‹ˆ μœ λ‚˜λŠ” μ•žμ—μ„œ 4번째, λ’€μ—μ„œ 6λ²ˆμ§Έμ— μžˆμ—ˆμŠ΅λ‹ˆλ‹€. μœ λ‚˜μ™€ μΉœκ΅¬λ“€μ€ λͺ¨λ‘ λͺ‡ λͺ…μž…λ‹ˆκΉŒ?
Comparison
var_a = 240 var_b = 2 var_c = var_a - var_b var_d = 2 var_e = var_c / var_d var_f = 2 var_g = var_e + var_f print(int(var_g))
121
121
240 2 [OP_SUB] 2 [OP_DIV] 2 [OP_ADD]
var_a = 240 var_b = 2 var_c = var_a - var_b var_d = 2 var_e = var_c / var_d var_f = 2 var_g = var_e + var_f print(int(var_g))
240 2 [OP_SUB] 2 [OP_DIV] 2 [OP_ADD]
There are 240 pencils and erasers in the pencil case. If there are 2 fewer pencils than erasers, how many erasers are there in the pencil case?
필톡에 μ—°ν•„κ³Ό μ§€μš°κ°œκ°€ λͺ¨λ‘ 240개 μžˆμŠ΅λ‹ˆλ‹€. 연필이 μ§€μš°κ°œλ³΄λ‹€ 2개 적게 λ“€μ–΄μžˆλ‹€λ©΄ 필톡 μ†μ˜ μ§€μš°κ°œλŠ” λͺ¨λ‘ λͺ‡ κ°œμž…λ‹ˆκΉŒ?
Arithmetic calculation
var_a = 38 var_b = 114 var_c = 152 var_d = 95 list_a= [] if "/" in str(var_d): var_d = eval(str(var_d)) list_a.append(var_d) if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_e = 1 list_b=list_a.copy() list_b.sort() var_f = list_b[-var_e] var_g = 1 list_c=list_a.copy() list_c.sort() var_h = list_c[var_g-1] var_i = var_f / var_h print(int(var_i))
4
4
[OP_LIST_SOL] 38 114 152 95 [OP_LIST_EOL] 1 [OP_LIST_MAX] 1 [OP_LIST_MIN] [OP_DIV]
var_a = 38 var_b = 114 var_c = 152 var_d = 95 list_a= [] if "/" in str(var_d): var_d = eval(str(var_d)) list_a.append(var_d) if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_e = 1 list_b=list_a.copy() list_b.sort() var_f = list_b[-var_e] var_g = 1 list_c=list_a.copy() list_c.sort() var_h = list_c[var_g-1] var_i = var_f / var_h print(int(var_i))
[OP_LIST_SOL] 38 114 152 95 [OP_LIST_EOL] 1 [OP_LIST_MAX] 1 [OP_LIST_MIN] [OP_DIV]
There are numbers which are 38, 114, 152, and 95. What is the value when the largest number is divided by the smallest number?
38, 114, 152, 95κ°€ μžˆμŠ΅λ‹ˆλ‹€. κ·Έμ€‘μ—μ„œ κ°€μž₯ 큰 수λ₯Ό κ°€μž₯ μž‘μ€ 수둜 λ‚˜λˆˆ 값은 μ–Όλ§ˆμž…λ‹ˆκΉŒ?
Arithmetic calculation
var_a = 20 var_b = 15 var_c = 18 var_d = 17 var_e = 20 list_a= [] if "/" in str(var_e): var_e = eval(str(var_e)) list_a.append(var_e) if "/" in str(var_d): var_d = eval(str(var_d)) list_a.append(var_d) if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_f = 18 list_b = [] for i in list_a: if i >= var_f: list_b.append(i) list_b = [float(i) for i in list_b] var_g = sum(list_b) print(int(var_g))
58
58
[OP_LIST_SOL] 20 15 18 17 20 [OP_LIST_EOL] 18 [OP_LIST_MORE_EQUAL] [OP_LIST_SUM]
var_a = 20 var_b = 15 var_c = 18 var_d = 17 var_e = 20 list_a= [] if "/" in str(var_e): var_e = eval(str(var_e)) list_a.append(var_e) if "/" in str(var_d): var_d = eval(str(var_d)) list_a.append(var_d) if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_f = 18 list_b = [] for i in list_a: if i >= var_f: list_b.append(i) list_b = [float(i) for i in list_b] var_g = sum(list_b) print(int(var_g))
[OP_LIST_SOL] 20 15 18 17 20 [OP_LIST_EOL] 18 [OP_LIST_MORE_EQUAL] [OP_LIST_SUM]
5 different boxes contain 20, 15, 18, 17, and 20 chocolates respectively. If all the chocolates are taken out of boxes containing 18 or more chocolates and put into one large box, find how many chocolates will fit in this box.
5개의 μ„œλ‘œ λ‹€λ₯Έ μƒμžμ— 초콜릿이 각각 20개, 15개, 18개, 17개, 20개 λ“€μ–΄μžˆμŠ΅λ‹ˆλ‹€. 초콜릿이 18개 이상 λ“€μ–΄μžˆλŠ” μƒμžλ“€μ—μ„œ μ΄ˆμ½œλ¦Ώμ„ λͺ¨λ‘ κΊΌλ‚΄ 큰 μƒμž ν•˜λ‚˜μ— λ‹΄μ•˜μ„ λ•Œ, 이 μƒμžμ—λŠ” λͺ‡ 개의 초콜릿이 λ‹΄κΈΈμ§€ κ΅¬ν•˜μ‹œμ˜€.
Comparison
var_a = 50 var_b = 5 var_c = 2 var_d = var_b * var_c var_e = var_a - var_d var_f = 1 var_g = 4 var_h = var_f + var_g var_i = var_e / var_h var_j = 4 var_k = var_i * var_j print(int(var_k))
32
32
50 5 2 [OP_MUL] [OP_SUB] 1 4 [OP_ADD] [OP_DIV] 4 [OP_MUL]
var_a = 50 var_b = 5 var_c = 2 var_d = var_b * var_c var_e = var_a - var_d var_f = 1 var_g = 4 var_h = var_f + var_g var_i = var_e / var_h var_j = 4 var_k = var_i * var_j print(int(var_k))
50 5 2 [OP_MUL] [OP_SUB] 1 4 [OP_ADD] [OP_DIV] 4 [OP_MUL]
Father's age divided by daughter's age equals 4. When the sum of their ages after 5 years is 50, how old is your father this year?
μ•„λ²„μ§€μ˜ λ‚˜μ΄λ₯Ό λ”Έμ˜ λ‚˜μ΄λ‘œ λ‚˜λˆ„λ©΄ 4μž…λ‹ˆλ‹€. 5λ…„ ν›„ 두 μ‚¬λžŒμ˜ λ‚˜μ΄λ₯Ό ν•©ν•˜λ©΄ 50일 λ•Œ, μ˜¬ν•΄ μ•„λ²„μ§€λŠ” λͺ‡ μ„Έμž…λ‹ˆκΉŒ?
Arithmetic calculation
var_a = 18 var_b = 12 var_c = var_a + var_b var_d = 2 var_e = var_c * var_d print(int(var_e))
60
60
18 12 [OP_ADD] 2 [OP_MUL]
var_a = 18 var_b = 12 var_c = var_a + var_b var_d = 2 var_e = var_c * var_d print(int(var_e))
18 12 [OP_ADD] 2 [OP_MUL]
The length of each side of the parallelogram is 18 centimeters (cm) and 12 centimeters (cm), respectively. Find the circumference of this parellelogram.
ν‰ν–‰μ‚¬λ³€ν˜•μ˜ 각 λ³€μ˜ κΈΈμ΄λŠ” 각각 18μ„Όν‹°λ―Έν„°(㎝), 12μ„Όν‹°λ―Έν„°(㎝)이닀. λ‘˜λ ˆλ₯Ό κ΅¬ν•˜λΌ.
Geometry
var_a = 30 var_b = 2 var_c = var_a / var_b var_d = 1 var_e = var_c - var_d var_f = 9 var_g = var_e * var_f print(int(var_g))
126
126
30 2 [OP_DIV] 1 [OP_SUB] 9 [OP_MUL]
var_a = 30 var_b = 2 var_c = var_a / var_b var_d = 1 var_e = var_c - var_d var_f = 9 var_g = var_e * var_f print(int(var_g))
30 2 [OP_DIV] 1 [OP_SUB] 9 [OP_MUL]
A total of 30 trees are planted 9 meters (m) apart on either side of a road from beginning to end. Find the length of this road in meters (m).
λ„λ‘œμ˜ μ–‘μͺ½μ— μ²˜μŒλΆ€ν„° λκΉŒμ§€ 총 30그루의 λ‚˜λ¬΄κ°€ 9λ―Έν„°(m) κ°„κ²©μœΌλ‘œ 심어져 μžˆμŠ΅λ‹ˆλ‹€. 이 λ„λ‘œμ˜ κΈΈμ΄λŠ” λͺ‡ λ―Έν„°(m)인지 ꡬ해 λ³΄μ‹œμ˜€.
Arithmetic calculation
var_a = 'A' var_b = 'B' var_c = 'C' list_a= [] if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_d = 1 var_e = 2 var_f = var_d / var_e var_g = 9 var_h = 10 var_i = var_g / var_h var_j = 2 var_k = 5 var_l = var_j / var_k list_b= [] if "/" in str(var_l): var_l = eval(str(var_l)) list_b.append(var_l) if "/" in str(var_i): var_i = eval(str(var_i)) list_b.append(var_i) if "/" in str(var_f): var_f = eval(str(var_f)) list_b.append(var_f) list_b.reverse() var_m = 1 list_c=list_b.copy() list_c.sort() var_n = list_c[var_m-1] var_o = list_b.index(var_n)+1 var_p = list_a[var_o-1] print(var_p)
C
C
[OP_LIST_SOL] A B C [OP_LIST_EOL] [OP_LIST_SOL] 1 2 [OP_DIV] 9 10 [OP_DIV] 2 5 [OP_DIV] [OP_LIST_EOL] 1 [OP_LIST_MIN] [OP_LIST_INDEX] [OP_LIST_POP] [OP_LIST_GET]
var_a = 'A' var_b = 'B' var_c = 'C' list_a= [] if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_d = 1 var_e = 2 var_f = var_d / var_e var_g = 9 var_h = 10 var_i = var_g / var_h var_j = 2 var_k = 5 var_l = var_j / var_k list_b= [] if "/" in str(var_l): var_l = eval(str(var_l)) list_b.append(var_l) if "/" in str(var_i): var_i = eval(str(var_i)) list_b.append(var_i) if "/" in str(var_f): var_f = eval(str(var_f)) list_b.append(var_f) list_b.reverse() var_m = 1 list_c=list_b.copy() list_c.sort() var_n = list_c[var_m-1] var_o = list_b.index(var_n)+1 var_p = list_a[var_o-1] print(var_p)
[OP_LIST_SOL] A B C [OP_LIST_EOL] [OP_LIST_SOL] 1 2 [OP_DIV] 9 10 [OP_DIV] 2 5 [OP_DIV] [OP_LIST_EOL] 1 [OP_LIST_MIN] [OP_LIST_INDEX] [OP_LIST_POP] [OP_LIST_GET]
There are three numbers A, B, and C. A is 1/2 and B is 9/10. If C is 2/5, what is the smallest among A, B, or C?
μ„Έ 개의 숫자 A, B, Cκ°€ μžˆμŠ΅λ‹ˆλ‹€. AλŠ” 1/2이고 BλŠ” 9/10μž…λ‹ˆλ‹€. Cκ°€ 2/5이면 A, B, C 쀑 μž‘μ€ μˆ˜λŠ” λ¬΄μ—‡μž…λ‹ˆκΉŒ?
Comparison
var_a = 36 var_b = 42 var_c = var_a + var_b var_d = 48 var_e = var_c + var_d var_f = 1 var_g = 0.16666666666666666 var_h = var_f - var_g var_i = var_e * var_h var_j = 97 var_k = var_i - var_j var_l = 3 var_m = 1 var_n = var_l - var_m var_o = var_k / var_n print(int(var_o))
4
4
36 42 [OP_ADD] 48 [OP_ADD] 1 1/6 [OP_SUB] [OP_MUL] 97 [OP_SUB] 3 1 [OP_SUB] [OP_DIV]
var_a = 36 var_b = 42 var_c = var_a + var_b var_d = 48 var_e = var_c + var_d var_f = 1 var_g = 0.16666666666666666 var_h = var_f - var_g var_i = var_e * var_h var_j = 97 var_k = var_i - var_j var_l = 3 var_m = 1 var_n = var_l - var_m var_o = var_k / var_n print(int(var_o))
36 42 [OP_ADD] 48 [OP_ADD] 1 1/6 [OP_SUB] [OP_MUL] 97 [OP_SUB] 3 1 [OP_SUB] [OP_DIV]
There are planks that are 36 centimeters (cm), 42 centimeters (cm), and 48 centimeters (cm) respectively. After cutting out 1/6 of this plank while trimming it, I overlapped and glued certain parts together, and the total length is 97 centimeters (cm). If all overlapping parts are the same length, find how many centimeters (cm) is the overlapping part.
길이가 각각 36μ„Όν‹°λ―Έν„°(㎝), 42μ„Όν‹°λ―Έν„°(㎝), 48μ„Όν‹°λ―Έν„°(㎝) 인 νŒμžκ°€ μžˆλ‹€. 이 판자λ₯Ό λ‹€λ“¬λŠλΌ 1/6을 μž˜λΌλ‚Έ ν›„ 일정 뢀뢄을 κ²Ήμ³μ„œ λΆ™ν˜”λ”λ‹ˆ 총 길이가 97μ„Όν‹°λ―Έν„°(㎝)이닀. κ²Ήμ²˜μ§„ λΆ€λΆ„μ˜ κΈΈμ΄λŠ” λͺ¨λ‘ κ°™μ„λ•Œ, κ²Ήμ²˜μ§„ 뢀뢄은 λͺ‡ μ„Όν‹°λ―Έν„°(㎝)인지 κ΅¬ν•˜μ‹œμ˜€.
Geometry
var_a = 43 var_b = 5 var_c = var_a - var_b var_d = 5 var_e = var_c * var_d print(int(var_e))
190
190
43 5 [OP_SUB] 5 [OP_MUL]
var_a = 43 var_b = 5 var_c = var_a - var_b var_d = 5 var_e = var_c * var_d print(int(var_e))
43 5 [OP_SUB] 5 [OP_MUL]
A number needs to be multiplied by 5, but I mistakenly added 5 to get 43. Find the result of the correct calculation.
μ–΄λ–€ μˆ˜μ— 5λ₯Ό κ³±ν•΄μ•Ό ν•˜λŠ”λ° 잘λͺ»ν•˜μ—¬ 5λ₯Ό λ”ν–ˆλ”λ‹ˆ 43이 λ˜μ—ˆμŠ΅λ‹ˆλ‹€. λ°”λ₯΄κ²Œ κ³„μ‚°ν•œ κ²°κ³Όλ₯Ό κ΅¬ν•˜μ‹œμ˜€.
Correspondence
var_a = 28 var_b = 4 var_c = var_a / var_b print(int(var_c))
7
7
28 4 [OP_DIV]
var_a = 28 var_b = 4 var_c = var_a / var_b print(int(var_c))
28 4 [OP_DIV]
You want to plant trees at intervals of 4 meters (m) in a circle-shaped park with a circumference of 28 meters (m). Find how many trees you can plant.
λ‘˜λ ˆκ°€ 28 λ―Έν„°(m) 인 원 λͺ¨μ–‘μ˜ 곡원에 4 λ―Έν„°(m) κ°„κ²©μœΌλ‘œ λ‚˜λ¬΄λ₯Ό μ‹¬μœΌλ €κ³  ν•©λ‹ˆλ‹€. λ‚˜λ¬΄λ₯Ό λͺ‡ 그루 심을 수 μžˆλŠ” μ§€ κ΅¬ν•˜μ‹œμ˜€.
Geometry
var_a = 50 var_b = 2 var_c = var_a - var_b var_d = 2 var_e = 3 var_f = var_d + var_e var_g = 1 var_h = var_f + var_g var_i = var_c / var_h var_j = 2 var_k = var_i * var_j print(int(var_k))
16
16
50 2 [OP_SUB] 2 3 [OP_ADD] 1 [OP_ADD] [OP_DIV] 2 [OP_MUL]
var_a = 50 var_b = 2 var_c = var_a - var_b var_d = 2 var_e = 3 var_f = var_d + var_e var_g = 1 var_h = var_f + var_g var_i = var_c / var_h var_j = 2 var_k = var_i * var_j print(int(var_k))
50 2 [OP_SUB] 2 3 [OP_ADD] 1 [OP_ADD] [OP_DIV] 2 [OP_MUL]
There is a prism whose sum of the number of edges, faces, and vertices is 50. How many edges of the pyramid have the same base shape as the prism?
λͺ¨μ„œλ¦¬, λ©΄, κΌ­μ§“μ μ˜ 수의 합이 50개인 각기λ‘₯이 μžˆλ‹€. 이 각기λ‘₯κ³Ό λ°‘λ©΄μ˜ λͺ¨μ–‘이 같은 κ°λΏ”μ˜ λͺ¨μ„œλ¦¬μ˜ μˆ˜λŠ” λͺ‡κ°œμΈκ°€μš”?
Geometry
var_a = 0 var_b = 1 var_c = 2 var_d = 5 list_a= [] if "/" in str(var_d): var_d = eval(str(var_d)) list_a.append(var_d) if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_e = 2 list_b = [str(i) for i in list_a] list_b = list(itertools.permutations(list_b, var_e)) list_b = [''.join(num_list) for num_list in list_b] list_b = [str_num for str_num in list_b if str_num[0] != '0'] list_b = [float(i) for i in list_b] var_f = 10 list_c = [] for i in list_b: if i > var_f: list_c.append(i) var_g = 100 list_d = [] for i in list_c: if i < var_g: list_d.append(i) var_h = len(list_d) print(int(var_h))
8
8
[OP_LIST_SOL] 0 1 2 5 [OP_LIST_EOL] 2 [OP_LIST_GET_PERM] 10 [OP_LIST_MORE] 100 [OP_LIST_LESS] [OP_LIST_LEN]
var_a = 0 var_b = 1 var_c = 2 var_d = 5 list_a= [] if "/" in str(var_d): var_d = eval(str(var_d)) list_a.append(var_d) if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_e = 2 list_b = [str(i) for i in list_a] list_b = list(itertools.permutations(list_b, var_e)) list_b = [''.join(num_list) for num_list in list_b] list_b = [str_num for str_num in list_b if str_num[0] != '0'] list_b = [float(i) for i in list_b] var_f = 10 list_c = [] for i in list_b: if i > var_f: list_c.append(i) var_g = 100 list_d = [] for i in list_c: if i < var_g: list_d.append(i) var_h = len(list_d) print(int(var_h))
[OP_LIST_SOL] 0 1 2 5 [OP_LIST_EOL] 2 [OP_LIST_GET_PERM] 10 [OP_LIST_MORE] 100 [OP_LIST_LESS] [OP_LIST_LEN]
You want to draw two numbers from the digits 0, 1, 2, and 5 to create a number greater than 10 and less than 100. In how many cases does each digit of the drawn number differ?
0, 1, 2, 5의 숫자 쀑 두 개의 숫자λ₯Ό 뽑아 10보닀 크고 100보닀 μž‘μ€ 수λ₯Ό λ§Œλ“€κ³ μž ν•œλ‹€. 뽑은 수 쀑 각 자리 μˆ«μžκ°€ λ‹€λ₯Έ κ²½μš°λŠ” λͺ‡ 가지인가?
Possibility
var_a = 36 var_b = 3 var_c = var_a - var_b var_d = 1 var_e = var_c - var_d var_f = 2 var_g = var_e / var_f var_h = 2 var_i = var_g ** var_h var_j = 36 var_k = var_i + var_j print(int(var_k))
292
292
36 3 [OP_SUB] 1 [OP_SUB] 2 [OP_DIV] 2 [OP_POW] 36 [OP_ADD]
var_a = 36 var_b = 3 var_c = var_a - var_b var_d = 1 var_e = var_c - var_d var_f = 2 var_g = var_e / var_f var_h = 2 var_i = var_g ** var_h var_j = 36 var_k = var_i + var_j print(int(var_k))
36 3 [OP_SUB] 1 [OP_SUB] 2 [OP_DIV] 2 [OP_POW] 36 [OP_ADD]
There are several ttakjis. I laid out the ttakjis to make a large square, and there were residual 36 ttakjis that were not used to make the square. I increased one horizontal and one vertical side by one column, and there are still three residual ttakjis left. How many ttakjis are there in all?
λ”±μ§€ μ—¬λŸ¬μž₯이 μžˆμŠ΅λ‹ˆλ‹€. λ”±μ§€λ₯Ό λŠ˜μ—¬λ†“μ•„ 큰 μ •μ‚¬κ°ν˜•μ„ λ§Œλ“€κ³  λ³΄λ‹ˆ 36μž₯의 λ”±μ§€κ°€ λ‚¨μ•˜μŠ΅λ‹ˆλ‹€. κ°€λ‘œ ν•œ λ³€κ³Ό μ„Έλ‘œ ν•œ 변을 1μ—΄μ”© λŠ˜λ Έλ”λ‹ˆ 아직도 3μž₯의 λ”±μ§€κ°€ λ‚¨μ•˜μŠ΅λ‹ˆλ‹€. λ”±μ§€λŠ” λͺ¨λ‘ λͺ‡ μž₯μž…λ‹ˆκΉŒ?
Geometry
var_a = 6 var_b = 8 var_c = var_a * var_b var_d = 4 var_e = var_c + var_d var_f = 4 var_g = var_e / var_f print(int(var_g))
13
13
6 8 [OP_MUL] 4 [OP_ADD] 4 [OP_DIV]
var_a = 6 var_b = 8 var_c = var_a * var_b var_d = 4 var_e = var_c + var_d var_f = 4 var_g = var_e / var_f print(int(var_g))
6 8 [OP_MUL] 4 [OP_ADD] 4 [OP_DIV]
When a number is multiplied by 4 and divided by 8, the quotient is 6 and the remainder is 4. Find out what this number is.
μ–΄λ–€ μˆ˜μ— 4λ₯Ό κ³±ν•œ 값을 8둜 λ‚˜λˆ„μ—ˆλ”λ‹ˆ λͺ«μ΄ 6이고 λ‚˜λ¨Έμ§€κ°€ 4μ˜€λ‹€. μ–΄λ–€ μˆ˜λŠ” μ–Όλ§ˆμΈμ§€ κ³ λ₯΄μ‹œμ˜€.
Correspondence
var_a = 1000 var_b = 8 var_c = var_a / var_b var_d = 0.3333333333333333 var_e = var_c ** var_d print(int(eval('{:.2f}'.format(round(var_e+1e-10,2)))))
5
5
1000 8 [OP_DIV] 1/3 [OP_POW]
var_a = 1000 var_b = 8 var_c = var_a / var_b var_d = 0.3333333333333333 var_e = var_c ** var_d print(int(eval('{:.2f}'.format(round(var_e+1e-10,2)))))
1000 8 [OP_DIV] 1/3 [OP_POW]
A cube is made by stacking 8 smaller cubes. If the volume of the stacked cubes is 1000 cubic centimeters (cm3), find the length of one corner of the smaller cube.
μ–΄λ–€ μ •μœ‘λ©΄μ²΄λŠ” 크기가 μž‘μ€ μ •μœ‘λ©΄μ²΄ 8개λ₯Ό μŒ“μ•„μ„œ λ§Œλ“€μ–΄μ‘ŒμŠ΅λ‹ˆλ‹€. μŒ“μ•„ λ§Œλ“€μ–΄μ§„ μ •μœ‘λ©΄μ²΄μ˜ λΆ€ν”Όκ°€ 1000μ„Έμ œκ³±μ„Όν‹°λ―Έν„°(㎀)일 λ•Œ, μž‘μ€ μ •μœ‘λ©΄μ²΄μ˜ ν•œ λͺ¨μ„œλ¦¬μ˜ 길이λ₯Ό κ΅¬ν•˜μ‹œμ˜€.
Geometry
var_a = 3 var_b = 10 var_c = var_a * var_b var_d = 9 var_e = var_c - var_d var_f = 3 var_g = var_e - var_f var_h = 10 var_i = 1 var_j = var_h - var_i var_k = var_g / var_j var_l = 3 var_m = 10 var_n = var_l * var_m var_o = var_k + var_n print(int(var_o))
32
32
3 10 [OP_MUL] 9 [OP_SUB] 3 [OP_SUB] 10 1 [OP_SUB] [OP_DIV] 3 10 [OP_MUL] [OP_ADD]
var_a = 3 var_b = 10 var_c = var_a * var_b var_d = 9 var_e = var_c - var_d var_f = 3 var_g = var_e - var_f var_h = 10 var_i = 1 var_j = var_h - var_i var_k = var_g / var_j var_l = 3 var_m = 10 var_n = var_l * var_m var_o = var_k + var_n print(int(var_o))
3 10 [OP_MUL] 9 [OP_SUB] 3 [OP_SUB] 10 1 [OP_SUB] [OP_DIV] 3 10 [OP_MUL] [OP_ADD]
There is a two-digit natural number whose tens digit is 3. If the number of digits in the tens digit and the ones digit of this natural number that are interchanged is 9 less than the initial number, find the initial number.
μ‹­μ˜ 자리의 μˆ«μžκ°€ 3인 두 자리 μžμ—°μˆ˜κ°€ μžˆλ‹€. 이 μžμ—°μˆ˜μ—μ„œ μ‹­μ˜ 자리의 μˆ«μžμ™€ 일의 자리의 숫자λ₯Ό μ„œλ‘œ λ°”κΎΌ μˆ˜κ°€ 처음 μˆ˜λ³΄λ‹€ 9만큼 μž‘λ‹€κ³  ν•  λ•Œ, 처음 수λ₯Ό κ΅¬ν•˜μ‹œμ˜€.
Arithmetic calculation
var_a = 1 var_b = 100 var_c = 1 list_a = [i for i in range(var_a, var_b + 1, var_c)] var_d = 7 var_e = 2 list_b = [] var_d = int(var_d) var_e = int(var_e) if var_e < 0: var_e = var_e + var_d for i in list_a: i = int(i) if i%var_d == var_e: list_b.append(i) var_f = 8 var_g = 2 list_c = [] var_f = int(var_f) var_g = int(var_g) if var_g < 0: var_g = var_g + var_f for i in list_b: i = int(i) if i%var_f == var_g: list_c.append(i) var_h = 1 list_d=list_c.copy() list_d.sort() var_i = list_d[var_h-1] print(int(var_i))
2
2
1 100 1 [OP_LIST_ARANGE] 7 2 [OP_LIST_DIVIDE_AND_REMAIN] 8 2 [OP_LIST_DIVIDE_AND_REMAIN] 1 [OP_LIST_MIN]
var_a = 1 var_b = 100 var_c = 1 list_a = [i for i in range(var_a, var_b + 1, var_c)] var_d = 7 var_e = 2 list_b = [] var_d = int(var_d) var_e = int(var_e) if var_e < 0: var_e = var_e + var_d for i in list_a: i = int(i) if i%var_d == var_e: list_b.append(i) var_f = 8 var_g = 2 list_c = [] var_f = int(var_f) var_g = int(var_g) if var_g < 0: var_g = var_g + var_f for i in list_b: i = int(i) if i%var_f == var_g: list_c.append(i) var_h = 1 list_d=list_c.copy() list_d.sort() var_i = list_d[var_h-1] print(int(var_i))
1 100 1 [OP_LIST_ARANGE] 7 2 [OP_LIST_DIVIDE_AND_REMAIN] 8 2 [OP_LIST_DIVIDE_AND_REMAIN] 1 [OP_LIST_MIN]
Among the numbers from 1 to 100, find the smallest number whose remainder when divided by 7 is 2 and when divided by 8 is also 2.
1λΆ€ν„° 100κΉŒμ§€μ˜ 수 쀑 7둜 λ‚˜λˆˆ λ‚˜λ¨Έμ§€κ°€ 2이고 8둜 λ‚˜λˆˆ λ‚˜λ¨Έμ§€λ„ 2인 수 쀑 κ°€μž₯ μž‘μ€ 값을 κ΅¬ν•˜μ‹œμ˜€.
Arithmetic calculation
var_a = 'AB+21=58' var_b = 'A' ans_dict = dict() var_a = var_a.replace('Γ—','*') var_a = var_a.replace('x','*') var_a = var_a.replace('Γ·','/') variable_candi = set(['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z']) for v in set(var_a): if v in variable_candi: ans_dict[v] = 1 candi = list(itertools.product('0123456789', repeat=len(ans_dict))) for c in candi: temp = var_a for i, (k, _) in enumerate(ans_dict.items()): temp = temp.replace(k, str(c[i])) term_list = [] op_list = [] temp_c = '' for tc in temp: if tc not in '+-*/=><().': temp_c += tc else: op_list.append(tc) term_list.append(temp_c) temp_c = '' term_list.append(temp_c) new_eq = '' for i in range(len(op_list)): new_eq += str(int(term_list[i]))+op_list[i] new_eq += str(int(term_list[-1])) if len(new_eq) == len(var_a): new_eq=new_eq.replace('=', '==') new_eq=new_eq.replace('>==', '>=') new_eq=new_eq.replace('<==', '<=') eval_result = False try: eval_result = eval(new_eq) except: pass if eval_result: for i, (k, _) in enumerate(ans_dict.items()): ans_dict[k] = int(c[i]) var_c = ans_dict[var_b] print(int(var_c))
3
3
AB+21=58 A [OP_DIGIT_UNK_SOLVER]
var_a = 'AB+21=58' var_b = 'A' ans_dict = dict() var_a = var_a.replace('Γ—','*') var_a = var_a.replace('x','*') var_a = var_a.replace('Γ·','/') variable_candi = set(['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z']) for v in set(var_a): if v in variable_candi: ans_dict[v] = 1 candi = list(itertools.product('0123456789', repeat=len(ans_dict))) for c in candi: temp = var_a for i, (k, _) in enumerate(ans_dict.items()): temp = temp.replace(k, str(c[i])) term_list = [] op_list = [] temp_c = '' for tc in temp: if tc not in '+-*/=><().': temp_c += tc else: op_list.append(tc) term_list.append(temp_c) temp_c = '' term_list.append(temp_c) new_eq = '' for i in range(len(op_list)): new_eq += str(int(term_list[i]))+op_list[i] new_eq += str(int(term_list[-1])) if len(new_eq) == len(var_a): new_eq=new_eq.replace('=', '==') new_eq=new_eq.replace('>==', '>=') new_eq=new_eq.replace('<==', '<=') eval_result = False try: eval_result = eval(new_eq) except: pass if eval_result: for i, (k, _) in enumerate(ans_dict.items()): ans_dict[k] = int(c[i]) var_c = ans_dict[var_b] print(int(var_c))
AB+21=58 A [OP_DIGIT_UNK_SOLVER]
AB+21=58. Calculate the value of A.
AB+21=58μž…λ‹ˆλ‹€. AλŠ” μ–Όλ§ˆμΈκ°€μš”?
Correspondence
var_a = 4 var_b = 6 var_c = 4 list_a= [] if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_d = 1 list_b=list_a.copy() list_b.sort() var_e = list_b[var_d-1] var_f = 1 list_c=list_a.copy() list_c.sort() var_g = list_c[var_f-1] var_h = 3 var_i = var_g ** var_h print(int(var_i))
64
64
[OP_LIST_SOL] 4 6 4 [OP_LIST_EOL] 1 [OP_LIST_MIN] 1 [OP_LIST_MIN] 3 [OP_POW]
var_a = 4 var_b = 6 var_c = 4 list_a= [] if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_d = 1 list_b=list_a.copy() list_b.sort() var_e = list_b[var_d-1] var_f = 1 list_c=list_a.copy() list_c.sort() var_g = list_c[var_f-1] var_h = 3 var_i = var_g ** var_h print(int(var_i))
[OP_LIST_SOL] 4 6 4 [OP_LIST_EOL] 1 [OP_LIST_MIN] 1 [OP_LIST_MIN] 3 [OP_POW]
There is a soap in the shape of a rectangular parallelepiped measuring 4 centimeters (cm) wide, 6 centimeters (cm) long, and 5 centimeters (cm) high. If you use this soap and it becomes a cube, what is the maximum volume of the soap?
κ°€λ‘œ 4μ„Όν‹°λ―Έν„°(㎝), μ„Έλ‘œ 6μ„Όν‹°λ―Έν„°(㎝), 높이 5μ„Όν‹°λ―Έν„°(㎝)인 직윑면체의 λͺ¨μ–‘μ˜ λΉ„λˆ„κ°€ μžˆμŠ΅λ‹ˆλ‹€. 이 λΉ„λˆ„λ₯Ό μ‚¬μš©ν•˜μ˜€λ”λ‹ˆ μ •μœ‘λ©΄μ²΄κ°€ λ˜μ—ˆλ‹€λ©΄, μ΄λ•Œ λΉ„λˆ„ λΆ€ν”Όμ˜ μ΅œλŒ€λŠ” μ–Όλ§ˆμž…λ‹ˆκΉŒ?
Geometry
var_a = 5 var_b = 11 var_c = var_a * var_b var_d = 43 var_e = var_c - var_d var_f = 5 var_g = 1 var_h = var_f - var_g var_i = var_e / var_h print(int(var_i))
3
3
5 11 [OP_MUL] 43 [OP_SUB] 5 1 [OP_SUB] [OP_DIV]
var_a = 5 var_b = 11 var_c = var_a * var_b var_d = 43 var_e = var_c - var_d var_f = 5 var_g = 1 var_h = var_f - var_g var_i = var_e / var_h print(int(var_i))
5 11 [OP_MUL] 43 [OP_SUB] 5 1 [OP_SUB] [OP_DIV]
This year the mother is 43 years old and the daughter is 11 years old. How many years ago was the mother's age five times her daughter's age?
μ˜¬ν•΄ μ–΄λ¨Έλ‹ˆλŠ” 43살이고 딸은 11μ‚΄μž…λ‹ˆλ‹€. μ–΄λ¨Έλ‹ˆ λ‚˜μ΄κ°€ λ”Έμ˜ λ‚˜μ΄μ˜ 5λ°°μ˜€μ„ λ•ŒλŠ” λͺ‡ λ…„ μ „μž…λ‹ˆκΉŒ?
Arithmetic calculation
var_a = 3 print(int(var_a))
3
3
3
var_a = 3 print(int(var_a))
3
How many edges are there on a face of an octahedron?
μ •νŒ”λ©΄μ²΄μ˜ ν•œ λ©΄μ—μ„œ 변은 λͺ‡ κ°œμΌκΉŒμš”?
Possibility
var_a = '7AB' ans_dict = dict() var_a = str(var_a) list_a = [] variable_candi = set(['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z']) for v in set(var_a): if v in variable_candi: ans_dict[v] = 0 candi = list(itertools.product('0123456789', repeat=len(ans_dict))) for c in candi: temp = var_a for i, (k, _) in enumerate(ans_dict.items()): temp = temp.replace(k, str(c[i])) if len(var_a) == len(str(int(temp))): new_elem = int(temp) list_a.append(new_elem) var_b = 7 list_b = [] var_b = int(var_b) for i in list_a: i = int(i) if i % var_b == 0: list_b.append(i) var_c = len(list_b) print(int(var_c))
15
15
7AB [OP_GEN_POSSIBLE_LIST] 7 [OP_LIST_DIVISIBLE] [OP_LIST_LEN]
var_a = '7AB' ans_dict = dict() var_a = str(var_a) list_a = [] variable_candi = set(['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z']) for v in set(var_a): if v in variable_candi: ans_dict[v] = 0 candi = list(itertools.product('0123456789', repeat=len(ans_dict))) for c in candi: temp = var_a for i, (k, _) in enumerate(ans_dict.items()): temp = temp.replace(k, str(c[i])) if len(var_a) == len(str(int(temp))): new_elem = int(temp) list_a.append(new_elem) var_b = 7 list_b = [] var_b = int(var_b) for i in list_a: i = int(i) if i % var_b == 0: list_b.append(i) var_c = len(list_b) print(int(var_c))
7AB [OP_GEN_POSSIBLE_LIST] 7 [OP_LIST_DIVISIBLE] [OP_LIST_LEN]
How many multiples of 7 are three-digit numbers with 7 in the hundreds place?
백의 μžλ¦¬κ°€ 7인 μ„Έ 자리수 쀑 7의 λ°°μˆ˜λŠ” λͺ‡ κ°œμž…λ‹ˆκΉŒ?
Correspondence
var_a = 66 var_b = 23 var_c = 2 var_d = var_b * var_c var_e = var_a + var_d var_f = 3 var_g = var_e * var_f var_h = 17 var_i = 23 var_j = var_h + var_i var_k = 2 var_l = var_j * var_k var_m = var_g - var_l var_n = 66 var_o = 2 var_p = var_n * var_o var_q = var_m - var_p var_r = 2 var_s = 2 var_t = var_r + var_s var_u = 3 var_v = var_t - var_u var_w = var_q / var_v print(int(var_w))
124
124
66 23 2 [OP_MUL] [OP_ADD] 3 [OP_MUL] 17 23 [OP_ADD] 2 [OP_MUL] [OP_SUB] 66 2 [OP_MUL] [OP_SUB] 2 2 [OP_ADD] 3 [OP_SUB] [OP_DIV]
var_a = 66 var_b = 23 var_c = 2 var_d = var_b * var_c var_e = var_a + var_d var_f = 3 var_g = var_e * var_f var_h = 17 var_i = 23 var_j = var_h + var_i var_k = 2 var_l = var_j * var_k var_m = var_g - var_l var_n = 66 var_o = 2 var_p = var_n * var_o var_q = var_m - var_p var_r = 2 var_s = 2 var_t = var_r + var_s var_u = 3 var_v = var_t - var_u var_w = var_q / var_v print(int(var_w))
66 23 2 [OP_MUL] [OP_ADD] 3 [OP_MUL] 17 23 [OP_ADD] 2 [OP_MUL] [OP_SUB] 66 2 [OP_MUL] [OP_SUB] 2 2 [OP_ADD] 3 [OP_SUB] [OP_DIV]
Three people each have a number card. If X is the average of the number cards of each person, the average of the number cards of two people A and B is 23 greater than X, and the average of the number cards of two people B and C is 17 less than X. The number written on B's number card was 66. Find the number written on A's number card.
μ„Έ μ‚¬λžŒμ΄ 숫자 μΉ΄λ“œλ₯Ό 각각 κ°€μ§€κ³  μžˆλ‹€. 각 μ‚¬λžŒμ΄ κ°€μ§„ 숫자 μΉ΄λ“œμ˜ 평균을 X라고 ν• λ•Œ, A, B 두 μ‚¬λžŒμ˜ 숫자 μΉ΄λ“œμ˜ 평균은 X보닀 23 크고, B, C λ‘μ‚¬λžŒμ˜ 숫자 평균은 X보닀 17μž‘λ‹€. B의 숫자 μΉ΄λ“œμ— 적힌 μˆ˜λŠ” 66μ΄μ—ˆλ‹€. A의 숫자 μΉ΄λ“œμ— 적힌 수λ₯Ό κ΅¬ν•˜μ—¬λΌ.
Comparison
var_a = 13 var_b = 4 var_c = var_a * var_b print(int(var_c))
52
52
13 4 [OP_MUL]
var_a = 13 var_b = 4 var_c = var_a * var_b print(int(var_c))
13 4 [OP_MUL]
A rabbit is walking in a park with a side length measuring 13 meters (m). How many meters (m) did the rabbit walk?
ν† λΌλŠ” ν•œ λ³€μ˜ 길이가 13λ―Έν„°(m)인 곡원을 μ‚°μ±…ν•˜κ³  μžˆμŠ΅λ‹ˆλ‹€. 토끼가 걸은 κ±°λ¦¬λŠ” 총 λͺ‡ λ―Έν„°(m)μž…λ‹ˆκΉŒ?
Geometry
var_a = 10 var_b = 1 var_c = var_a - var_b print(int(var_c))
9
9
10 1 [OP_SUB]
var_a = 10 var_b = 1 var_c = var_a - var_b print(int(var_c))
10 1 [OP_SUB]
When 10 is divided by A, the quotient is B and the remainder is C. A, B, and C are natural numbers. If B and C are equal, find the largest possible number of A.
10λ₯Ό A둜 λ‚˜λˆ΄μ„ λ•Œ λͺ«μ΄ B, λ‚˜λ¨Έμ§€κ°€ C μ˜€μŠ΅λ‹ˆλ‹€. A, B, CλŠ” μžμ—°μˆ˜ μž…λ‹ˆλ‹€. B와 Cκ°€ 같을 λ•Œ, κ°€λŠ₯ν•œ A 쀑 κ°€μž₯ 큰 수λ₯Ό κ΅¬ν•˜μ‹œμ˜€.
Correspondence
var_a = 45 var_b = 15 var_c = var_a / var_b var_d = 1 var_e = var_c - var_d print(int(var_e))
2
2
45 15 [OP_DIV] 1 [OP_SUB]
var_a = 45 var_b = 15 var_c = var_a / var_b var_d = 1 var_e = var_c - var_d print(int(var_e))
45 15 [OP_DIV] 1 [OP_SUB]
Yoongi multiplied 15 by a certain number to get 45. Find the exact result of subtracting 1 from the number.
μœ€κΈ°λŠ” 15λ₯Ό μ–΄λ–€ μˆ˜μ— κ³±ν•΄ 45λ₯Ό μ–»μ—ˆμŠ΅λ‹ˆλ‹€. μ–΄λ–€ μˆ˜μ—μ„œ 1λ₯Ό λΊ€ κ²°κ³Όλ₯Ό μ •ν™•νžˆ κ΅¬ν•˜μ‹œμ˜€.
Correspondence
var_a = 3 var_b = 3 var_c = 4 var_d = var_b + var_c var_e = var_a + var_d print(int(var_e))
10
10
3 3 4 [OP_ADD] [OP_ADD]
var_a = 3 var_b = 3 var_c = 4 var_d = var_b + var_c var_e = var_a + var_d print(int(var_e))
3 3 4 [OP_ADD] [OP_ADD]
Jungkook has 3 marbles, Jimin has 4 more marbles than what Jungkook has. What is a total number of their marbles?
μ •κ΅­μ΄λŠ” κ΅¬μŠ¬μ„ 3개 κ°€μ§€κ³  있고, 지민이가 κ°€μ§„ κ΅¬μŠ¬μ€ 정ꡭ이보닀 4κ°œκ°€ 더 λ§ŽμŠ΅λ‹ˆλ‹€. 두 μ‚¬λžŒμ΄ κ°€μ§€κ³  μžˆλŠ” κ΅¬μŠ¬μ€ λͺ¨λ‘ λͺ‡ κ°œμž…λ‹ˆκΉŒ?
Arithmetic calculation
var_a = 120 var_b = 6 var_c = 2 var_d = var_b * var_c var_e = var_a / var_d print(int(var_e))
10
10
120 6 2 [OP_MUL] [OP_DIV]
var_a = 120 var_b = 6 var_c = 2 var_d = var_b * var_c var_e = var_a / var_d print(int(var_e))
120 6 2 [OP_MUL] [OP_DIV]
The sum of the edges of a hexagonal pyramid is 120 centimeters (cm). What is the sum of the edges divided by the number of edges?
μ–΄λ–€ μœ‘κ°λΏ”μ˜ λͺ¨μ„œλ¦¬μ˜ 합은 120μ„Όν‹°λ―Έν„°(㎝)μž…λ‹ˆλ‹€. λͺ¨μ„œλ¦¬μ˜ 합을 λͺ¨μ„œλ¦¬μ˜ 개수둜 λ‚˜λˆ„λ©΄ μ–Όλ§ˆμž…λ‹ˆκΉŒ?
Geometry
var_a = '31+6AB=683' var_b = 'B' ans_dict = dict() var_a = var_a.replace('Γ—','*') var_a = var_a.replace('x','*') var_a = var_a.replace('Γ·','/') variable_candi = set(['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z']) for v in set(var_a): if v in variable_candi: ans_dict[v] = 1 candi = list(itertools.product('0123456789', repeat=len(ans_dict))) for c in candi: temp = var_a for i, (k, _) in enumerate(ans_dict.items()): temp = temp.replace(k, str(c[i])) term_list = [] op_list = [] temp_c = '' for tc in temp: if tc not in '+-*/=><().': temp_c += tc else: op_list.append(tc) term_list.append(temp_c) temp_c = '' term_list.append(temp_c) new_eq = '' for i in range(len(op_list)): new_eq += str(int(term_list[i]))+op_list[i] new_eq += str(int(term_list[-1])) if len(new_eq) == len(var_a): new_eq=new_eq.replace('=', '==') new_eq=new_eq.replace('>==', '>=') new_eq=new_eq.replace('<==', '<=') eval_result = False try: eval_result = eval(new_eq) except: pass if eval_result: for i, (k, _) in enumerate(ans_dict.items()): ans_dict[k] = int(c[i]) var_c = ans_dict[var_b] print(int(var_c))
2
2
31+6AB=683 B [OP_DIGIT_UNK_SOLVER]
var_a = '31+6AB=683' var_b = 'B' ans_dict = dict() var_a = var_a.replace('Γ—','*') var_a = var_a.replace('x','*') var_a = var_a.replace('Γ·','/') variable_candi = set(['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z']) for v in set(var_a): if v in variable_candi: ans_dict[v] = 1 candi = list(itertools.product('0123456789', repeat=len(ans_dict))) for c in candi: temp = var_a for i, (k, _) in enumerate(ans_dict.items()): temp = temp.replace(k, str(c[i])) term_list = [] op_list = [] temp_c = '' for tc in temp: if tc not in '+-*/=><().': temp_c += tc else: op_list.append(tc) term_list.append(temp_c) temp_c = '' term_list.append(temp_c) new_eq = '' for i in range(len(op_list)): new_eq += str(int(term_list[i]))+op_list[i] new_eq += str(int(term_list[-1])) if len(new_eq) == len(var_a): new_eq=new_eq.replace('=', '==') new_eq=new_eq.replace('>==', '>=') new_eq=new_eq.replace('<==', '<=') eval_result = False try: eval_result = eval(new_eq) except: pass if eval_result: for i, (k, _) in enumerate(ans_dict.items()): ans_dict[k] = int(c[i]) var_c = ans_dict[var_b] print(int(var_c))
31+6AB=683 B [OP_DIGIT_UNK_SOLVER]
We have the expression 31+6AB=683. Find the number that goes into B
식 31+6AB=683이 μžˆμŠ΅λ‹ˆλ‹€. B에 λ“€μ–΄κ°€μ•Ό ν•˜λŠ” 숫자λ₯Ό κ΅¬ν•˜μ‹œμ˜€
Correspondence
var_a = 'μ •κ΅­' var_b = '윀기' list_a= [] if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_c = 6 var_d = 3 var_e = var_c - var_d var_f = 4 list_b= [] if "/" in str(var_f): var_f = eval(str(var_f)) list_b.append(var_f) if "/" in str(var_e): var_e = eval(str(var_e)) list_b.append(var_e) list_b.reverse() var_g = 1 list_c=list_b.copy() list_c.sort() var_h = list_c[-var_g] var_i = list_b.index(var_h)+1 var_j = list_a[var_i-1] print(var_j)
윀기
Yoongi
[OP_LIST_SOL] μ •κ΅­ 윀기 [OP_LIST_EOL] [OP_LIST_SOL] 6 3 [OP_SUB] 4 [OP_LIST_EOL] 1 [OP_LIST_MAX] [OP_LIST_INDEX] [OP_LIST_POP] [OP_LIST_GET]
var_a = 'Jungkook' var_b = 'Yoongi' list_a= [] if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_c = 6 var_d = 3 var_e = var_c - var_d var_f = 4 list_b= [] if "/" in str(var_f): var_f = eval(str(var_f)) list_b.append(var_f) if "/" in str(var_e): var_e = eval(str(var_e)) list_b.append(var_e) list_b.reverse() var_g = 1 list_c=list_b.copy() list_c.sort() var_h = list_c[-var_g] var_i = list_b.index(var_h)+1 var_j = list_a[var_i-1] print(var_j)
[OP_LIST_SOL] Jungkook Yoongi [OP_LIST_EOL] [OP_LIST_SOL] 6 3 [OP_SUB] 4 [OP_LIST_EOL] 1 [OP_LIST_MAX] [OP_LIST_INDEX] [OP_LIST_POP] [OP_LIST_GET]
Jungkook ate 3 out of 6 apples. If Yoongi has 4 apples, who has more apples?
μ •κ΅­μ΄λŠ” 6개의 μ‚¬κ³Όμ—μ„œ 3개λ₯Ό λ¨Ήμ—ˆμŠ΅λ‹ˆλ‹€. μœ€κΈ°κ°€ κ°€μ§„ 사과가 4개라면 λˆ„κ°€ 더 많이 사과λ₯Ό κ°€μ§€κ³  μžˆλ‚˜μš”?
Comparison
var_a = 7 var_b = 2 var_c = 9 list_a= [] if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_d = 3 list_b = [str(i) for i in list_a] list_b = list(itertools.permutations(list_b, var_d)) list_b = [''.join(num_list) for num_list in list_b] list_b = [str_num for str_num in list_b if str_num[0] != '0'] list_b = [float(i) for i in list_b] var_e = len(list_b) print(int(var_e))
6
6
[OP_LIST_SOL] 7 2 9 [OP_LIST_EOL] 3 [OP_LIST_GET_PERM] [OP_LIST_LEN]
var_a = 7 var_b = 2 var_c = 9 list_a= [] if "/" in str(var_c): var_c = eval(str(var_c)) list_a.append(var_c) if "/" in str(var_b): var_b = eval(str(var_b)) list_a.append(var_b) if "/" in str(var_a): var_a = eval(str(var_a)) list_a.append(var_a) list_a.reverse() var_d = 3 list_b = [str(i) for i in list_a] list_b = list(itertools.permutations(list_b, var_d)) list_b = [''.join(num_list) for num_list in list_b] list_b = [str_num for str_num in list_b if str_num[0] != '0'] list_b = [float(i) for i in list_b] var_e = len(list_b) print(int(var_e))
[OP_LIST_SOL] 7 2 9 [OP_LIST_EOL] 3 [OP_LIST_GET_PERM] [OP_LIST_LEN]
You want to create a three-digit number using the numbers 7, 2, and 9 only once. How many three-digit numbers can you make?
숫자 7, 2, 9λ₯Ό ν•œ λ²ˆμ”©λ§Œ μ‚¬μš©ν•˜μ—¬ μ„Έ 자리 수λ₯Ό λ§Œλ“€λ €κ³  ν•©λ‹ˆλ‹€. λ§Œλ“€ 수 μžˆλŠ” μ„Έ 자리 μˆ˜λŠ” λͺ¨λ‘ λͺ‡ κ°œμž…λ‹ˆκΉŒ?
Possibility
var_a = 1650 var_b = 50 var_c = var_a / var_b var_d = 1 var_e = var_c + var_d var_f = 2 var_g = var_e * var_f print(int(var_g))
68
68
1650 50 [OP_DIV] 1 [OP_ADD] 2 [OP_MUL]
var_a = 1650 var_b = 50 var_c = var_a / var_b var_d = 1 var_e = var_c + var_d var_f = 2 var_g = var_e * var_f print(int(var_g))
1650 50 [OP_DIV] 1 [OP_ADD] 2 [OP_MUL]
At intervals of 50 meters (m), statues are erected on both sides of the street a distance of 1650 meters (m). If there are statues at the beginning and end of the street, how many statues are there on both sides of the street?
50λ―Έν„°(m) κ°„κ²©μœΌλ‘œ 동상이 1650λ―Έν„°(m) κ±°λ¦¬μœ„μ— μ„Έμ›Œμ ΈμžˆμŠ΅λ‹ˆλ‹€. 거리의 처음과 끝에도 동상이 μ„Έμ›Œμ Έ μžˆμ„ λ•Œ, 거리 μ–‘μͺ½μ— μ„Έμ›Œμ§„ λ™μƒμ˜ μˆ˜λŠ” λͺ¨λ‘ λͺ‡ 개 μž…λ‹ˆκΉŒ?
Arithmetic calculation
var_a = 3 var_b = 12 var_c = var_a * var_b var_d = 8 var_e = var_c - var_d var_f = 17 var_g = var_e - var_f print(int(var_g))
11
11
3 12 [OP_MUL] 8 [OP_SUB] 17 [OP_SUB]
var_a = 3 var_b = 12 var_c = var_a * var_b var_d = 8 var_e = var_c - var_d var_f = 17 var_g = var_e - var_f print(int(var_g))
3 12 [OP_MUL] 8 [OP_SUB] 17 [OP_SUB]
Jimin had three dozen pencils. He gave 8 of them to his younger brother and a few to Yuna, so he has 17 left. If one dozen is 12 pencils, how many pencils did Jimin give Yuna?
μ§€λ―Όμ΄λŠ” μ—°ν•„ 3타λ₯Ό κ°€μ§€κ³  μžˆμ—ˆμŠ΅λ‹ˆλ‹€. κ·Έ 쀑 8자루λ₯Ό λ™μƒμ—κ²Œ μ£Όκ³  μœ λ‚˜μ—κ²Œλ„ λͺ‡ 자루 μ£Όμ—ˆλ”λ‹ˆ 17μžλ£¨κ°€ λ‚¨μ•˜μŠ΅λ‹ˆλ‹€. μ—°ν•„ 1νƒ€λŠ” 12μžλ£¨μΌλ•Œ, 지민이가 μœ λ‚˜μ—κ²Œ μ€€ 연필은 λͺ‡ μžλ£¨μΌκΉŒμš”?
Arithmetic calculation
var_a = 30 var_b = 3 var_c = var_a / var_b var_d = 1 var_e = var_c - var_d print(int(var_e))
9
9
30 3 [OP_DIV] 1 [OP_SUB]
var_a = 30 var_b = 3 var_c = var_a / var_b var_d = 1 var_e = var_c - var_d print(int(var_e))
30 3 [OP_DIV] 1 [OP_SUB]
If the sum of three consecutive natural numbers is 30, find the smallest natural number of them.
μ—°μ†ν•˜λŠ” μ„Έ μžμ—°μˆ˜μ˜ 합이 30일 λ•Œ, μ„Έ μžμ—°μˆ˜ 쀑 첫번째 수λ₯Ό κ΅¬ν•˜μ‹œμ˜€.
Arithmetic calculation
var_a = 5 var_b = 3 var_c = 1 var_a = int(var_a) var_b = int(var_b) for i, elem in enumerate(range(var_b)): var_c = var_c * (var_a-i) for i, elem in enumerate(range(var_b)): var_c = var_c / (i+1) print(int(var_c))
10
10
5 3 [OP_COMB]
var_a = 5 var_b = 3 var_c = 1 var_a = int(var_a) var_b = int(var_b) for i, elem in enumerate(range(var_b)): var_c = var_c * (var_a-i) for i, elem in enumerate(range(var_b)): var_c = var_c / (i+1) print(int(var_c))
5 3 [OP_COMB]
Yoongi wants to buy 3 out of 5 books at the bookstore. Find how many possible cases there are.
μœ€κΈ°λŠ” μ„œμ μ—μ„œ 5개의 μ±… 쀑 3개의 책을 사렀고 ν•©λ‹ˆλ‹€. κ°€λŠ₯ν•œ 경우의 μˆ˜λŠ” λͺ¨λ‘ λͺ‡ 가지인지 κ΅¬ν•˜μ‹œμ˜€.
Possibility