11
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Make the result $-132,680$ using these operations and numbers:

Numbers (cannot be used more than once, but not all need to be used):

$2,4,6,8,9,10,11,12,23,27$

Operations (all operations have to be used once and once only):

  • $\Box + \Box$
  • $\Box - \Box$
  • $\Box \times \Box$
  • $\Box \div \Box$
  • $\Box^{\Box}$
  • $\log_{10}{\Box}$
  • $\sqrt[2]{\Box}$
  • $\Box !$

Hints:

The ! operation needs to be done first

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7
  • 1
    $\begingroup$ Can you clarify whether the log and root functions are unary or binary and, if unary, what base the former has? $\endgroup$ – msh210 May 10 '20 at 22:10
  • $\begingroup$ @msh210 Base 10 $\endgroup$ – Xnero May 10 '20 at 22:11
  • $\begingroup$ @msh210 Unary, the root has a base of 2 (square-root) of a single number $\endgroup$ – Xnero May 10 '20 at 22:21
  • $\begingroup$ How can there be 10 numbers but only 5 binary operators? Is there implicit multiplication/division or concatenation of numbers? $\endgroup$ – DenverCoder1 May 11 '20 at 9:18
  • $\begingroup$ @eyl327 I should make it clear all numbers do not need to be used $\endgroup$ – Xnero May 11 '20 at 9:20
6
+50
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This is what I've found:

$$ \frac{6! + \log\left(10\right) - 27 ^ 4}{\sqrt{2 \times 8}} = -132680 $$

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1
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    $\begingroup$ Well done, that's correct $\endgroup$ – Xnero May 16 '20 at 10:57
9
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Okay here's one way to do it (each operation used once).

$$-132680 = \left(\frac{6}{\log(\sqrt{10})} \times (8^2 + 4) \right) - !9$$ where $!9$ is the subfactorial of $9$ ($!9 = 133496$)

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    $\begingroup$ Not my intended method but well done anyway $\endgroup$ – Xnero May 12 '20 at 23:43
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    $\begingroup$ Ah rot13(Fhosnpgbevny!) That's really neat. I didn't expect that interpretation. +1 $\endgroup$ – Galen May 13 '20 at 0:52

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