Make -132,680 from these operations

Question

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

Can you clarify whether the log and root functions are unary or binary and, if unary, what base the former has? — msh210, Commented May 10, 2020 at 22:10
@msh210 Unary, the root has a base of 2 (square-root) of a single number — user68905, Commented May 10, 2020 at 22:21
How can there be 10 numbers but only 5 binary operators? Is there implicit multiplication/division or concatenation of numbers? — DenverCoder1, Commented May 11, 2020 at 9:18
@eyl327 I should make it clear all numbers do not need to be used — user68905, Commented May 11, 2020 at 9:20

cyberpotato · Accepted Answer · 2020-05-16 10:54:30Z

5

+50

This is what I've found:

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

answered May 16, 2020 at 10:54

cyberpotato

2311 silver badge5 bronze badges

1

$\begingroup$ Well done, that's correct $\endgroup$
– user68905
Commented May 16, 2020 at 10:57

Add a comment |

hexomino · Accepted Answer · 2020-05-12 23:40:18Z

8

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 $9$ ($!9 = 133496$)

answered May 12, 2020 at 23:40

hexomino

139k10 gold badges397 silver badges576 bronze badges

1

$\begingroup$ Not my intended method but well done anyway $\endgroup$
– user68905
Commented May 12, 2020 at 23:43
1

$\begingroup$ Ah rot13(Fhosnpgbevny!) That's really neat. I didn't expect that interpretation. +1 $\endgroup$
– Galen
Commented May 13, 2020 at 0:52

Add a comment |

2 Answers 2