# Make -132,680 from these operations

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

This is what I've found:

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

• Well done, that's correct – Daniil May 16 at 10:57

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 $$9$$ ($$!9 = 133496$$)

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