philschmid/DMath huggingface dataset:
Category: Possibility

Question (EN): You want to create a six-digit number using 2, 0, 4, 1, 5, and 8 only once. What is the sum of the largest and smallest possible numbers?
Question (KO): 2, 0, 4, 1, 5, 8를 한 번씩만 사용하여 여섯 자리 수를 만드려고 합니다. 만들 수 있는 수 중에서 가장 큰 수와 가장 작은 수를 더하면 얼마가 됩니까?

Answer: 956668

Solution (Python):
var_a = 2
var_b = 0
var_c = 4
var_d = 1
var_e = 5
var_f = 8
list_a= []
if "/" in str(var_f):
    var_f = eval(str(var_f))
list_a.append(var_f)
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_g = 6
list_b = [str(i) for i in list_a]
list_b = list(itertools.permutations(list_b, var_g))
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_h = 1
list_c=list_b.copy()
list_c.sort()
var_i = list_c[-var_h]
var_j = 1
list_d=list_b.copy()
list_d.sort()
var_k = list_d[var_j-1]
var_l = var_i + var_k
print(int(var_l))
Category: Correspondence

Question (EN): There are two different natural numbers, A and B. When A is divided by 17, the quotient is 25 and the remainder is B. Find A when the remainder B is the largest.
Question (KO): 서로 다른 두 자연수 A, B가 있습니다. A를 17로 나누면 몫은 25이고 나머지는 B가 됩니다. 나머지 B가 가장 큰 수일 때 A를 구하시오.

Answer: 441

Solution (Python):
var_a = 25
var_b = 17
var_c = var_a * var_b
var_d = 17
var_e = 1
var_f = var_d - var_e
var_g = var_c + var_f
print(int(var_g))
Category: Arithmetic calculation

Question (EN): The sum of 7 consecutive natural numbers is 63. Find the middle number.
Question (KO): 연속하는 7개의 자연수의 합이 63입니다. 가운데 수를 구하시오.

Answer: 9

Solution (Python):
var_a = 63
var_b = 0
var_c = 7
var_d = 1
var_e = var_c - var_d
var_f = 1
list_a = [i for i in range(var_b, var_e + 1, var_f)]
list_a = [float(i) for i in list_a]
var_g = sum(list_a)
var_h = var_a - var_g
var_i = 7
var_j = var_h / var_i
var_k = 7
var_l = 2
var_m = var_k // var_l
var_n = var_j + var_m
print(int(var_n))
Category: Geometry

Question (EN): Jungkook has 5 red balls, 4 blue balls, and 3 yellow balls. If Yoon-gi gives Jungkook 1 yellow ball, how many blue balls does Jungkook have?
Question (KO): 정국이는 빨간 공 5개, 파란 공 4개, 노란 공 3개를 가지고 있습니다. 윤기가 정국이에게 노란 공 1개를 주었을 때, 정국이는 파란 공을 몇 개 가지고 있습니까?

Answer: 4

Solution (Python):
var_a = 4
print(int(var_a))
Category: Arithmetic calculation

Question (EN): What is the sum of all numbers that are divisible by 4 and 10 when you choose 3 cards with 1, 2, 4, and 0 written on them and use them all to make a three-digit number?
Question (KO): 1, 2, 4, 0가 적힌 카드 중 3장을 골라 모두 사용하여 세 자리 수를 만들 때, 4와 10으로 나누어 떨어지는 모든 수의 합은 얼마인가?

Answer: 920

Solution (Python):
var_a = 1
var_b = 2
var_c = 4
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 = 4
list_c = []
var_f = int(var_f)
for i in list_b:
    i = int(i)
    if i % var_f == 0:
        list_c.append(i)
var_g = 10
list_d = []
var_g = int(var_g)
for i in list_c:
    i = int(i)
    if i % var_g == 0:
        list_d.append(i)
list_d = [float(i) for i in list_d]
var_h = sum(list_d)
print(int(var_h))
Category: Correspondence

Question (EN): When you add 7 to a number, you get 27. Divide a number by 5, then add 5 and multiply by 7. What is the result?
Question (KO): 어떤 수에 7을 더하였더니 27이 되었습니다. 어떤 수를 5로 나눈 다음 5를 더하고 7을 곱한 값을 구하시오.

Answer: 63

Solution (Python):
var_a = 27
var_b = 7
var_c = var_a - var_b
var_d = 5
var_e = var_c / var_d
var_f = 5
var_g = var_e + var_f
var_h = 7
var_i = var_g * var_h
print(int(var_i))
Category: Comparison

Question (EN): Yoongi collected 7, Jungkook 6, and Yuna 9. Who has the smallest number?
Question (KO): 윤기는 7, 정국은 6, 유나는 9만큼 모았습니다. 가장 작은 숫자를 가진 사람은 누구입니까?

Answer: Jungkook

Solution (Python):
var_a = 'Yoongi'
var_b = 'Jungkook'
var_c = 'Yuna'
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 = 7
var_e = 6
var_f = 9
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-1]
var_i = list_b.index(var_h)+1
var_j = list_a[var_i-1]
print(var_j)
Category: Arithmetic calculation

Question (EN): Junhyeok bought 7 apple boxes with 16 apples in each box, and Jihyun bought 6 apple boxes with 25 apples in each box. How many apples did Junhyeok and Jihyun buy?
Question (KO): 준혁이는 한 상자에 16개의 사과가 들어있는 사과상자 7개를 샀고, 지현이는 한 상자에 25개의 사과가 들어있는 사과상자 6개를 샀습니다. 준혁이와 지현이가 산 사과는 모두 몇 개인가요?

Answer: 262

Solution (Python):
var_a = 16
var_b = 7
var_c = var_a * var_b
var_d = 25
var_e = 6
var_f = var_d * var_e
var_g = var_c + var_f
print(int(var_g))
Category: Comparison

Question (EN): When 43.23, 2/5, and 21.23 are arranged in order from smallest to greatest, which number comes last?
Question (KO): 43.23, 2/5 및 21.23을 작은 순서로 배열할 때, 마지막으로 오는 수는 무엇입니까?

Answer: 43.23

Solution (Python):
var_a = 43.23
var_b = 2
var_c = 5
var_d = var_b / var_c
var_e = 21.23
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_a):
    var_a = eval(str(var_a))
list_a.append(var_a)
list_a.reverse()
var_f = len(list_a)
list_b=list_a.copy()
list_b.sort()
var_g = list_b[var_f-1]
print('{:.2f}'.format(round(var_g+1e-10,2)))
Category: Geometry

Question (EN): You have a die with the sum of the number of eyes on one side and the number of eyes on the other side is 7. Given that the eyes on this die are 1 through 6, find the number of eyes on the opposite side of 5 multiplied by 5.
Question (KO): 마주 보는 면의 눈의 수의 합이 7인 주사위가 있습니다. 이 주사위에 그려진 눈은 1부터 6까지라고 할 때, 5와 마주보는 면의 눈의 수와 5를 곱한 값은 얼마인지 구하시오.

Answer: 10

Solution (Python):
var_a = 5
var_b = 7
var_c = 5
var_d = var_b - var_c
var_e = var_a * var_d
print(int(var_e))
Category: Correspondence

Question (EN): Given that 3.A7>3.27 is true, how many digits from 0 to 9 can be written for A?
Question (KO): 3.A7>3.27입니다. 0부터 9까지의 숫자 중에서 A에 쓸 수 있는 숫자는 모두 몇 개입니까?

Answer: 7

Solution (Python):
var_a = '3.A7>3.27'
var_b = 'A'
ans_dict = dict()
var_a = var_a.replace('×','*')
var_a = var_a.replace('x','*')
var_a = var_a.replace('÷','/')
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] = []
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].append(int(c[i]))
list_a = list(set(ans_dict[var_b]))
var_c = len(list_a)
print(int(var_c))
Category: Correspondence

Question (EN): There is a three-digit number A27 that can be made from a single-digit number A. Rounding down this number from the hundreds place gives 200. What is A?
Question (KO): 한 자리 수 A로 만들 수 있는 세 자리 수 A27이 있습니다. 이 수를 백의 자리에서 버림하면 200이 됩니다. A는 무엇입니까?

Answer: 2

Solution (Python):
var_a = 'A27'
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 = 200
list_b = []
for i in list_a:
    if i >= var_b:
        list_b.append(i)
var_c = 200
var_d = 100
var_e = var_c + var_d
list_c = []
for i in list_b:
    if i < var_e:
        list_c.append(i)
var_f = 'A'
var_a = str(var_a)
var_f = str(var_f)
unk_idx = var_a.index(var_f)
var_g = 0
for elem in list_c:
    elem = str(elem)
    var_g = int(elem[unk_idx])
print(int(var_g))
Category: Comparison

Question (EN): Seokjin is standing right in front of Jimin. If Jimin is 9th from the front, what number is Seokjin?
Question (KO): 석진이가 지민이 바로 앞에 서 있습니다. 지민이가 앞에서 9번째라면 석진이는 몇 번째입니까?

Answer: 8

Solution (Python):
var_a = 9
var_b = 1
var_c = var_a - var_b
print(int(var_c))
Category: Arithmetic calculation

Question (EN): Apples are on sale in packs of 4 at the supermarket. If you buy 2 bunches, how many apples are there in total?
Question (KO): 마트에서 사과를 4개 묶음으로 팝니다. 2 묶음을 샀다면 사과는 총 몇 개입니까?

Answer: 8

Solution (Python):
var_a = 4
var_b = 2
var_c = var_a * var_b
print(int(var_c))
Category: Geometry

Question (EN): If a rectangle with an area of 36.48 square centimeters (cm2) has a width of 6.08 centimeters (cm), how many centimeters (cm) is its length?
Question (KO): 넓이가 36.48제곱센티미터(㎠)인 직사각형의 가로가 6.08센티미터(㎝)라면 세로는 몇 센티미터(㎝)입니까?

Answer: 6

Solution (Python):
var_a = 36.48
var_b = 6.08
var_c = var_a / var_b
print(int(eval('{:.2f}'.format(round(var_c+1e-10,2)))))
Category: Correspondence

Question (EN): A is a single digit number. If you round down A76 in the hundreds place, you get a value of 700. What is A?
Question (KO): A는 한 자리 수입니다. A76을 백의 자리에서 버림하면 700의 값을 얻을 수 있습니다. A는 무엇입니까?

Answer: 7

Solution (Python):
var_a = 'A76'
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 = 700
list_b = []
for i in list_a:
    if i >= var_b:
        list_b.append(i)
var_c = 700
var_d = 100
var_e = var_c + var_d
list_c = []
for i in list_b:
    if i < var_e:
        list_c.append(i)
var_f = 'A76'
var_g = 'A'
var_f = str(var_f)
var_g = str(var_g)
unk_idx = var_f.index(var_g)
var_h = 0
for elem in list_c:
    elem = str(elem)
    var_h = int(elem[unk_idx])
print(int(var_h))
Category: Correspondence

Question (EN): A number that needs to be divided by 9/4, but multiplied by mistake, got 18/5. How much do you get when calculated correctly?
Question (KO): 어떤 수에 9/4를 나누어야 하는데 잘못하여 곱했더니 18/5가 되었습니다. 바르게 계산하면 얼마가 됩니까?

Answer: 0.71

Solution (Python):
var_a = 3.6
var_b = 2.25
var_c = var_a / var_b
var_d = 2.25
var_e = var_c / var_d
print('{:.2f}'.format(round(var_e+1e-10,2)))
Category: Arithmetic calculation

Question (EN): If there are 250 kilograms (kg) of carrots in the supermarket, and 2.5 times the weight of carrots is the weight of cucumbers, how many kilograms (kg) are there both carrots and cucumbers?
Question (KO): 마트에 당근 250킬로그램(㎏)이 있고, 당근 무게의 2.5배가 오이의 무게라면 당근과 오이는 모두 몇 킬로그램(㎏) 있습니까?

Answer: 875

Solution (Python):
var_a = 250
var_b = 250
var_c = 2.5
var_d = var_b * var_c
var_e = var_a + var_d
print(int(var_e))
Category: Comparison

Question (EN): sunplant was planted in 2/7 of the flower garden, and snapweed was planted in 3/8. Which flowers are more widely planted?
Question (KO): 꽃밭의 2/7에는 채송화 3/8에는 봉선화를 심었습니다. 어떤 꽃이 더 넓게 심어 져 있습니까?

Answer: snapweed

Solution (Python):
var_a = 'sunplant'
var_b = 'snapweed'
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 = 2
var_d = 7
var_e = var_c / var_d
var_f = 3
var_g = 8
var_h = var_f / var_g
list_b= []
if "/" in str(var_h):
    var_h = eval(str(var_h))
list_b.append(var_h)
if "/" in str(var_e):
    var_e = eval(str(var_e))
list_b.append(var_e)
list_b.reverse()
var_i = 1
list_c=list_b.copy()
list_c.sort()
var_j = list_c[-var_i]
var_k = list_b.index(var_j)+1
var_l = list_a[var_k-1]
print(var_l)
Category: Correspondence

Question (EN): When you divided a number by 1.5 incorrectly, which was supposed to be divided by 6, the quotient is 3.8. Find the answer if you calculate correctly.
Question (KO): 어떤 수를 6으로 나누어야 할 것을 잘못하여 1.5로 나누었더니 몫이 3.8이 되었습니다. 바르게 계산하면 얼마인지 답을 구해 보세요.

Answer: 0.95

Solution (Python):
var_a = 3.8
var_b = 1.5
var_c = var_a * var_b
var_d = 6
var_e = var_c / var_d
print('{:.2f}'.format(round(var_e+1e-10,2)))
Category: Possibility

Question (EN): How many different three-digit numbers can be formed using the three numbers 1, 2, and 7 once each?
Question (KO): 1, 2, 7을 각각 한번씩만 사용하여 세 자리 수를 만들면 몇 가지의 다른 수를 만들 수 있겠는가?

Answer: 6

Solution (Python):
var_a = 1
var_b = 2
var_c = 7
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))
Category: Possibility

Question (EN): How many common multiples of 6 and 9 are found when you pick 3 of the natural number cards in sequence from 0 to 4 to form a 3-digit number?
Question (KO): 0부터 4까지 차례대로 있는 자연수 카드 중 3장을 꺼내 세 자리 수를 만들 때, 6과 9의 공배수는 모두 몇 가지가 나오는가?

Answer: 4

Solution (Python):
var_a = 0
var_b = 1
var_c = 2
var_d = 3
var_e = 4
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 = 3
list_b = [str(i) for i in list_a]
list_b = list(itertools.permutations(list_b, var_f))
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_g = 6
var_h = 9
var_i = var_h * var_g / math.gcd(int(var_h), int(var_g))
list_c = []
var_i = int(var_i)
for i in list_b:
    i = int(i)
    if i % var_i == 0:
        list_c.append(i)
var_j = len(list_c)
print(int(var_j))
Category: Possibility

Question (EN): Colored pencils of the same color cannot be distinguished. Find the number of ways to take out 2 colored pencils at a time when there are 2 or more colored pencils in each of red, blue, green, and purple.
Question (KO): 같은 색의 색연필은 구별이 되지 않는다. 빨간색, 파란색, 초록색, 보라색 색연필이 각각 2자루 이상 통에 들어 있을때 한 번에 색연필 2자루를 꺼내는 경우의 수를 구하시오.

Answer: 6

Solution (Python):
var_a = 4
var_b = 3
var_c = var_a * var_b
var_d = 2
var_e = var_c / var_d
print(int(var_e))
Category: Comparison

Question (EN): Friends lined up in a row. Seokjin stands 8th from the front and 6th from the back. How many friends are all in line?
Question (KO): 친구들이 일렬로 줄을 서 있습니다. 앞에서 8번째, 뒤에서 6번째에 석진이가 서 있습니다. 줄을 서 있는 친구들은 모두 몇 명입니까?

Answer: 13

Solution (Python):
var_a = 8
var_b = 6
var_c = var_a + var_b
var_d = 1
var_e = var_c - var_d
print(int(var_e))
Category: Possibility

Question (EN): You are trying to create a three-digit number by choosing 3 of the 4 number cards 1, 3, 4, 7 and using them only once. How many numbers that are greater than or equal to 360 and less than or equal to 455 can be made?
Question (KO): 4 장의 숫자 카드 1, 3, 4, 7 중 3 장을 뽑아 한 번씩만 사용하여 세 자리 수를 만들려고 합니다. 만들 수 있는 수 중에서 360 이상 455 이하인 수는 모두 몇 개입니까?

Answer: 6

Solution (Python):
var_a = 1
var_b = 3
var_c = 4
var_d = 7
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 = 360
list_c = []
for i in list_b:
    if i >= var_f:
        list_c.append(i)
var_g = 455
list_d = []
for i in list_c:
    if i <= var_g:
        list_d.append(i)
var_h = len(list_d)
print(int(var_h))
Category: Geometry

Question (EN): There is a cube in the shape of a cube, and the sum of the lengths of all its edges is 72 centimeters (cm). Find the length in centimeters (cm) of one edge of this cube.
Question (KO): 모든 모서리의 길이의 합이 72센티미터(㎝)인 정육면체 모양의 큐브가 있습니다. 이 큐브의 한 모서리의 길이는 몇 센티미터(㎝)인지 구하시오.

Answer: 6

Solution (Python):
var_a = 72
var_b = 4
var_c = 3
var_d = var_b * var_c
var_e = var_a / var_d
print(int(var_e))
Category: Possibility

Question (EN): When two of the number cards 3, 7, and 2 are used once to create the smallest two-digit number, which number is unused?
Question (KO): 숫자 카드 3, 7, 2 중 2장을 한 번씩 모두 사용하여 가장 작은 두 자리 수를 만들 때, 사용하지 않는 카드의 숫자는 무엇입니까?

Answer: 7

Solution (Python):
var_a = 3
var_b = 7
var_c = 2
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 = [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 = 1
list_c=list_b.copy()
list_c.sort()
var_f = list_c[var_e-1]
list_d = []
var_f = int(var_f)
while var_f//10 > 0:
    list_d.append(var_f%10)
    var_f = var_f//10
list_d.append(var_f%10)
list_d = list_d[::-1]
list_e = list(set(list_a) - set(list_d))
var_g = 1
var_h = list_e[var_g-1]
print(int(var_h))
Category: Correspondence

Question (EN): B is 278 plus 365 three times, and A is a number consisting of 20 bundles of 100 and 87 bundles of 10. Find A+B.
Question (KO): B는 278에 365를 3번 더한 숫자이고 A는 100개 묶음 20개 10개 묶음 87개로 이루어진 숫자입니다. A+B를 구하시오.

Answer: 4243

Solution (Python):
var_a = 100
var_b = 20
var_c = var_a * var_b
var_d = 10
var_e = 87
var_f = var_d * var_e
var_g = var_c + var_f
var_h = 278
var_i = 365
var_j = 3
var_k = var_i * var_j
var_l = var_h + var_k
var_m = var_g + var_l
print(int(var_m))
Category: Geometry

Question (EN): After connecting 5 rice cakes with an original length of 2.7meters (m), overlapping each other by 0.3meters (m), cut them into 6 slices again to give them to the neighbors fairly. How much should I give?
Question (KO): 원래 길이가 2.7인 떡 5개를, 0.3미터(m)씩 겹치게 하여 이은 후, 다시 6도막으로 잘라 이웃 주민들에게 공평하게 주려고 합니다. 얼마씩 주어야 합니까?

Answer: 2.05

Solution (Python):
var_a = 2.7
var_b = 5
var_c = var_a * var_b
var_d = 0.3
var_e = 5
var_f = 1
var_g = var_e - var_f
var_h = var_d * var_g
var_i = var_c - var_h
var_j = 6
var_k = var_i / var_j
print('{:.2f}'.format(round(var_k+1e-10,2)))
Category: Arithmetic calculation

Question (EN): A 1000 milliliter (ml) bucket was filled with water, but there was a hole and water was leaking at a rate of 5 milliliters (ml) per minute. Find the time elapsed when 300 milliliters (ml) of water remain.
Question (KO): 1000밀리리터(㎖)짜리 물통에 물이 가득 차 있었는데, 구멍이 생겨 1분에 5밀리리터(㎖)씩 물이 새고 있다. 물이 300밀리리터(㎖) 남아있을 때까지 지난 시간을 구하시오.

Answer: 140

Solution (Python):
var_a = 1000
var_b = 300
var_c = var_a - var_b
var_d = 5
var_e = var_c / var_d
print(int(var_e))
Category: Possibility

Question (EN): When using 3 of the number cards 1, 4, 6, and 7 to make a 3-digit number, find the third smallest number with 1 in the tens place.
Question (KO): 숫자카드 1, 4, 6, 7 중 3개를 사용하여 3자리 수를 만들때, 십의 자리 수가 1인 세 번째 작은 수를 구하십시오.

Answer: 614

Solution (Python):
var_a = 1
var_b = 4
var_c = 6
var_d = 7
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 = 10
var_g = 1
list_c = [] 
var_f = int(var_f)
var_g = int(var_g)
for i in list_b:
    i = int(i)
    if (i//var_f)%10 == var_g:
        list_c.append(i)
var_h = 3
list_d=list_c.copy()
list_d.sort()
var_i = list_d[var_h-1]
print(int(var_i))
Category: Correspondence

Question (EN): When a certain number is divided by the number of students in Hyungjun's class, the quotient is 1.2, and when the number of students in Hyungjun's class is divided by the number of students in Jinmok's class, the quotient is 5/6. If you divide a certain number by the number of students in Jinmok's class, what is the quotient?
Question (KO): 어떤 수를 형준이네 반 학생 수로 나누면 몫이 1.2로 나누어떨어지고, 형준이네 반 학생수를 다시 진목이네 반 학생수로 나누면 몫이 5/6로 나누어떨어진다. 어떤 수를 진목이네 반 학생수로 나누면 그 몫은 얼마인가?

Answer: 1

Solution (Python):
var_a = 1.2
var_b = 0.8333333333333334
var_c = var_a * var_b
print(int(var_c))
Category: Geometry

Question (EN): In a rhombus with an area of 64/5 square centimeters (cm2), two non-adjacent vertices were selected and the distance was measured, which was 64/9 centimeters (cm). Find the distance between two unselected vertices.
Question (KO): 넓이가 64/5제곱센티미터(㎠)인 마름모에서 이웃하지 않은 두 꼭짓점을 선택해 거리를 재었더니 64/9센티미터(㎝)였습니다. 선택되지 않은 두 꼭짓점 사이의 거리를 구하시오.

Answer: 3.6

Solution (Python):
var_a = 12.8
var_b = 2
var_c = var_a * var_b
var_d = 7.111111111111111
var_e = var_c / var_d
print('{:.2f}'.format(round(var_e+1e-10,2)))
Category: Correspondence

Question (EN): When you add 10 to a particular number and divide it by 5, you get 4. Multiply that particular number by 3 and subtract 18. What is the value?
Question (KO): 어떤 수에 10을 더하여 5로 나눈 수가 4입니다. 어떤 수에 3을 곱하여 18을 뺀 수를 구하시오.

Answer: 12

Solution (Python):
var_a = 4
var_b = 5
var_c = var_a * var_b
var_d = 10
var_e = var_c - var_d
var_f = 3
var_g = var_e * var_f
var_h = 18
var_i = var_g - var_h
print(int(var_i))