text: # Make $21$ out of $1,5,6,7$ (challenge).
  
  Here is a small mathematical challenge.
  
  You have to use once and only once each of the four numbers $1,5,6,7$ in order to obtain, via the help of the usual operators ($+,-,/,*$ together with parenthesis) in order to obtain the number $21$.
  
  Example of an unsuccessful computation: $6*5-7-1=22\neq 21$
  
  -
  $6 / (1 - 5/7)$ ${}$ – Cocopuffs Jun 12 '13 at 19:33
  @Cocopuffs Arghh you beat me! – Kaish Jun 12 '13 at 19:35
  Hint : Give this problem to a (non-)mathematician friend and bet that he can't solve it in 1 hour without googling it. You may often win... – user81566 Jun 12 '13 at 20:43
  
  #!/usr/bin/env python
  # -*- coding: utf-8 -*-
  from fractions import Fraction
  
  L=[Fraction(1),Fraction(5),Fraction(6),Fraction(7)]
  O={'+':10,'-':10,'*':10,'/':10}
  N=Fraction(21)
  
  P = {'+':lambda x,y:x+y,
  '-':lambda x,y:x-y,
  '*':lambda x,y:x*y,
  '/':lambda x,y:x/y}
  
  def postfix_search(rest, stk):
  if len(stk) >= 2:
  y = (v2, r2) = stk.pop()
  x = (v1, r1) = stk.pop()
  for opr in O:
  if O[opr] > 0 and not (opr == '/' and v2 == 0):
  stk += [(P[opr](v1, v2), '('+r1+opr+r2+')')]
  O[opr] -= 1
  if postfix_search(rest, stk): return 1
  O[opr] += 1
  stk.pop()
  stk += [x, y]
  elif not rest:
  v, r = stk[0]
  if v == N: print(r)
  return v == N
  for x in list(rest):
  rest.remove(x)
  stk += [x]
  if postfix_search(rest, stk):
  return True
  stk.pop()
  rest += [x]
  
  if __name__ == "__main__":
  postfix_search(list(zip(L, map(str, L))), [])
  
  
  Run
  
  -
explanation: The document presents a mathematical challenge to create the number 21 using the numbers 1, 5, 6, and 7 exactly once with the operations +, -, *, and /. It includes an example of an unsuccessful attempt and a successful solution provided in a comment. The document also contains a Python script that solves the problem programmatically.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Make $21$ out of $1,5,6,7$ (challenge).
  
  Here is a small mathematical challenge.
  
  You have to use once and only once each of the four numbers $1,5,6,7$ in order to obtain, via the help of the usual operators ($+,-,/,*$ together with parenthesis) in order to obtain the number $21$.
  
  Example of an unsuccessful computation: $6*5-7-1=22\neq 21$
  
  -
  $6 / (1 - 5/7)$ ${}$ – Cocopuffs Jun 12 '13 at 19:33
  @Cocopuffs Arghh you beat me! – Kaish Jun 12 '13 at 19:35
  Hint : Give this problem to a (non-)mathematician friend and bet that he can't solve it in 1 hour without googling it. You may often win... – user81566 Jun 12 '13 at 20:43
  
  #!/usr/bin/env python
  # -*- coding: utf-8 -*-
  from fractions import Fraction
  
  L=[Fraction(1),Fraction(5),Fraction(6),Fraction(7)]
  O={'+':10,'-':10,'*':10,'/':10}
  N=Fraction(21)
  
  P = {'+':lambda x,y:x+y,
  '-':lambda x,y:x-y,
  '*':lambda x,y:x*y,
  '/':lambda x,y:x/y}
  
  def postfix_search(rest, stk):
  if len(stk) >= 2:
  y = (v2, r2) = stk.pop()
  x = (v1, r1) = stk.pop()
  for opr in O:
  if O[opr] > 0 and not (opr == '/' and v2 == 0):
  stk += [(P[opr](v1, v2), '('+r1+opr+r2+')')]
  O[opr] -= 1
  if postfix_search(rest, stk): return 1
  O[opr] += 1
  stk.pop()
  stk += [x, y]
  elif not rest:
  v, r = stk[0]
  if v == N: print(r)
  return v == N
  for x in list(rest):
  rest.remove(x)
  stk += [x]
  if postfix_search(rest, stk):
  return True
  stk.pop()
  rest += [x]
  
  if __name__ == "__main__":
  postfix_search(list(zip(L, map(str, L))), [])
  
  
  Run
  
  -
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document presents a mathematical challenge to create the number 21 using the numbers 1, 5, 6, and 7 exactly once with the operations +, -, *, and /. It includes an example of an unsuccessful attempt and a successful solution provided in a comment. The document also contains a Python script that solves the problem programmatically.
  
  Label: question-and-answer