6
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As the visual says, get the right equation by using the numbers exactly once.

You must use ALL five numbers exactly once.

You cannot use any other math operators.

enter image description here

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    $\begingroup$ Could you provide a text version of the puzzle? Unless there's some sort of reason why the stylized image is necessary for the enigmatic character of the puzzle? $\endgroup$
    – bobble
    Jan 11 at 22:41
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    $\begingroup$ @bobble. See the answer. The numbers have to be presented in a particular way. $\endgroup$
    – RogerA
    Jan 12 at 14:11

4 Answers 4

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I can make

54÷6=9 or 54÷9=6

Wait... How?

You can overlap 1 and 7, and then rotate it 180 degrees to make 4.
image

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    $\begingroup$ Very Creative @ACB. Exactly what I had in mind $\endgroup$
    – RogerA
    Jan 12 at 14:09
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Rendering the equation as A / B = C we can make some observations:

  1. Solutions for B and C are interchangeable. That is, if there is a solution B=x and C=y then B=y and C=x is also a solution.
  2. A has exactly 2 digits.
  • If A had 3 digits then we would be left with 1-digit numbers for the divisors (B and C), which prevents the product (A) from being >= 100.
  • If A had 1 digit then either B or C would be greater than A which would not work.
  1. Consequently, B and C have 3 digits between them, sinnce we have exactly 5 digits to work with and A has 2 of them.
  2. Neither B nor C can be exactly 9. If either was 9, then the sum of digits of A would have to be a multiple of 9, and there is no combination of 2 of the remaining digits that meet this requirement.
  3. Neither B nor C can be exactly 1. If either was 1, then the other would have to be equal to A. But we can't re-use digits, so this is impossible.
  4. Neither B nor C can be exactly 5. If either was 5, then A would need to end in zero or 5. But we can't re-use 5 and zero is not available. (Similarly, neither can end in 5.)
  5. Neither B nor C can be exactly 6. If either was 6, then A would need to end with an even digit. But we have no even digits other than 6. (Similarly, neither can end in 6.)
  6. A can not end in 5 or 6, for the same reasons given in the previous two observations.
  7. Since either B or C must be a single digit (from (3)), and neither can be exactly 1, 5, 6, or 9, then one of them must be 7. Let's use B=7 (since solutions for B and C are symmetrical)
  8. we are now down to the following possibilities for A: 17,19,51,57,59,61,67,69,71,79,91,97,99 - of which only 91 is a multiple of 7. But 91 / 7 = 13, not 65 or 56.

So I conclude there is no solution "using all the numbers exactly once".

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It isn't the most creative of solutions, but there are contexts in which

97/15 = 6, for example, Integer Arithmetic

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If you permit answers in fields other than the reals, there are rather a lot of solutions*! For example, in the numbers mod 1697, 1 / 5 = 679, because 679 * 5 = 1697 * 2 + 1 ≡ 1 mod 1697.

* The exact number of solutions depends on some choices you must make about which, if any, of the question marks you allow to be larger than the modulus you are working in. If none are required to be normalized in this way, there are 1910 solutions, but even if you require all three question marks to be "small" there are still 1044 solutions.

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