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How do you make $21$ from the numbers $1$, $5$, $6$, and $7$?

  • You can use the operations of addition, subtraction, multiplication and division, as well as brackets.
  • You must use each number exactly once.
  • You cannot juxtapose numbers (i.e., 1 and 5 cannot be used as 15).
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    $\begingroup$ I would answer, but I found this link from another site that asks the same question. $\endgroup$ – Andrea Gottardi Nov 6 '14 at 15:59
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    $\begingroup$ meta.puzzling.stackexchange.com/questions/1348/… $\endgroup$ – nicael Nov 6 '14 at 16:00
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    $\begingroup$ I thought the issue was with copying copyrighted material verbatim, without permission or attribution. I don't see plagiarism here. $\endgroup$ – frodoskywalker Nov 6 '14 at 16:07
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    $\begingroup$ I am not talking specifically about this question, but the very fact that googling gives you answers cannot be a criterion for deletion is my contention $\endgroup$ – skv Nov 6 '14 at 16:17
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    $\begingroup$ If being able to Google a puzzle is reason for deletion, then probably the vast majority of questions on this site would have to be deleted. $\endgroup$ – pacoverflow Nov 6 '14 at 16:27
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If this is an acceptable question, here is the answer:

$$\frac{6}{1-\frac{5}{7}}=21$$

This is a pretty well known problem that, while the math obviously works, is unintuitive enough that it is sometimes difficult to solve the first time you see it.

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Observe

$1 = \frac{5}{7} + \frac{6}{21}$

and rearrange.

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I came across this question a long time ago and after struggling with it came up with this:

$ \binom {6 + 1}{7 - 5}$

which is

$\frac {7!}{2!5!}= \frac {7 \times 6}{2} = 21$

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