11
$\begingroup$

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

$\endgroup$
  • 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 at 22:10
  • $\begingroup$ @msh210 Base 10 $\endgroup$ – Daniil May 10 at 22:11
  • $\begingroup$ @msh210 Unary, the root has a base of 2 (square-root) of a single number $\endgroup$ – Daniil May 10 at 22:21
  • $\begingroup$ How can there be 10 numbers but only 5 binary operators? Is there implicit multiplication/division or concatenation of numbers? $\endgroup$ – eyl327 May 11 at 9:18
  • $\begingroup$ @eyl327 I should make it clear all numbers do not need to be used $\endgroup$ – Daniil May 11 at 9:20
6
+50
$\begingroup$

This is what I've found:

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

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Well done, that's correct $\endgroup$ – Daniil May 16 at 10:57
9
$\begingroup$

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 $9$ ($!9 = 133496$)

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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