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Create an equation using 1, 6, 7 and 8 to equal 2. Can use brackets, plus, minus, subtract, multiply.

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    $\begingroup$ Welcome to Puzzling, take our tour! Could you please provide proper attribution for this question? $\endgroup$
    – bobble
    Apr 21 at 16:34
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    $\begingroup$ Is concatenation allowed, and can a number be used more than once? $\endgroup$ Apr 21 at 16:36
  • $\begingroup$ Every number has to be used once. And only once. Thanks for your help! $\endgroup$ Apr 21 at 17:20
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    $\begingroup$ Thank you. You don't mention 'divide'. Is division allowed? And did you make this puzzle? $\endgroup$ Apr 21 at 17:29

3 Answers 3

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A silly answer:

If you write the numbers in Roman numerals
I VI VII VIII
and reorganise the letters
V - V + V - III + II - II
the sum is II or $2$.

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Partial answer for when division is allowed:

((8 + 6) / 7) x 1

Detailing that gives:

  1. 8 + 6 = 14

  2. 14 / 7 = 2

  3. 2 x 1 = 2

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impossible

1 6 7 8 mod 7 is 1 -1 0 1
so to make 2, we need 2 mod 7. We need to add or subtract at least once, and the addition has to involve something that is not 0 mod 7.

To make 2 mod 7, there is either 1 + 1 or 1 - -1 (or their negation). The other two numbers could then only make -1, 0, or 1 mod 7: you can either add 0 or multiply by +- 1.
1 + 1 --> 1 + 8 = 9
1 - -1 --> 1 - 6 = -5 or 8 - 6 = 2

In the 1 + 1 case, only 0 and -1 (7 and 6) are left.
Multiplying by +- 1 doesn't work, and making 0 is only possible by 7 * 6.

In the 1 - -1 case, 7 and either 1 or 8 are left.
Adding 0: 7 * 1 only works with 1 - 6, repeating 1 (the 7 * 1 can just be 7 but the 8 can't be placed anywhere in 7 + 1 - 6), 7 * 8 doesn't work
Multiplying by +- 1: 7 - 8 only works with 8 - 6, repeating 8

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