• You may use up to two subtraction symbols (no negation symbols).
  • You may use up to two decimal points.
  • You will use one vinculum for division (no "/").
  • You may use concatenation of digits only, including forming numbers which have decimal points in them.
  • You may not use grouping symbols such as parentheses, brackets, or braces.
  • You may not use factorial signs.
  • You may not use square roots.
  • You may not use exponentiation.
  • You may not use logarithms.
  • You may not use trigonometric functions.
  • You may not use any other characters or operations.
  • This is in base 10. (The numbers using the ones, and 1,000 on the other side of the equals sign, are in base 10.)

Try to create six different solutions.

  • $\begingroup$ To clarify are you allowed to use any 0's or exclusively the six 1's and decimals? $\endgroup$
    – gabbo1092
    Jan 30 at 17:42
  • 1
    $\begingroup$ @gabbo1092 -- No zeroes may be used. You can write .1, 1.1, .11, 1.11, etc., if they would make expressions have the correct value(s). $\endgroup$ Jan 30 at 18:37
  • $\begingroup$ bin-->dec(1111101000) $\endgroup$ Jan 31 at 10:03

1 Answer 1


Solution 1


Solution 2


Solution 3


Solution 4


Solution 5


Solution 6


  • 1
    $\begingroup$ 1) I would not consider the negative signs you used in solutions 2 and 4 to be subtraction symbols. 2) Along with the first point, and adding in solution 6, solutions 2, 4, and 6 are not of "six different solutions" in the spirit of the forms A/B = -A/(-B). $\endgroup$ Jan 30 at 18:43
  • $\begingroup$ @ hexomino -- Attention! Your solutions 3 and 4 equal -1,000. $\endgroup$ Jan 30 at 18:50
  • $\begingroup$ @OliveStemforn Apologies on 3 and 4, edited now. Sorry, also, my understanding of "subtraction symbol" was that negation was okay, my mistake. $\endgroup$
    – hexomino
    Jan 30 at 18:56
  • $\begingroup$ hexomino, you have essentially three different solutions out of the six different solutions in your solutions 1, 3, and 5. There are still three other solutions significantly different from those that I am looking for from any contributor(s). $\endgroup$ Jan 30 at 19:14
  • 4
    $\begingroup$ @OliveStemforn I've replaced 2, 4 and 6 now with something essentially different $\endgroup$
    – hexomino
    Jan 30 at 19:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.