You have to solve the following:

\(0\) *_ \(0\) _* \(0 = 6\)

\(1\) *_ \(1\) _* \(1 = 6\)

\(2\) *_ \(2\) _* \(2 = 6\)

\(3\) *_ \(3\) _* \(3 = 6\)

\(4\) *_ \(4\) _* \(4 = 6\)

\(5\) *_ \(5\) _* \(5 = 6\)

\(6\) *_ \(6\) _* \(6 = 6\)

\(7\) *_ \(7\) _* \(7 = 6\)

\(8\) *_ \(8\) _* \(8 = 6\)

\(9\) *_ \(9\) _* \(9 = 6\)

You can put as many operations in the spaces but they must not contain a digit/letter. e.g.

You can have \(2 \times \sqrt{2} + 2!\) but not \(2^2 + \sqrt [3]{2} + 2e\).

You can only use factorials and double factorials, BODMAS/PEMDAS operations and square roots.

Good luck.

**Extension**: Find the smallest positive integral value of \(n\) such that \(n\) *_ \(n\) _* \(n = 6\) has no solutions.

## Comments

Sort by:

TopNewestAnswer below. If you want to do it yourself, don't scroll down: \[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\]\[.\] \((0!+0!+0!)!\)

\((1+1+1)!\)

\(2+2+2\)

\(3!+3-3\)

\(\sqrt { 4 } +\sqrt { 4 } +\sqrt { 4 } \)

\((5\div5)+5\)

\(6+6-6\)

\(7-(7\div7)\)

\(\left( \sqrt { 8+8 } !\div 8 \right) !\)

\(\sqrt { 9 } +\sqrt { \sqrt { 9 } \times \sqrt { 9 } } \)

Legid? – Julian Poon · 1 year, 11 months ago

Log in to reply

\( 9 \)

_ \( 9\) _\(9 = 10\)Without addition. – Krishna Sharma · 1 year, 4 months ago

Log in to reply

– Julian Poon · 1 year, 4 months ago

\(-(-9\div9-9)\)Log in to reply

– Krishna Sharma · 1 year, 4 months ago

Sir, you are genius!Log in to reply

– Julian Poon · 1 year, 4 months ago

But \(--\) is \(+\)...Log in to reply

– Sharky Kesa · 1 year, 4 months ago

Try the extension.Log in to reply

– Julian Poon · 1 year, 4 months ago

After the crazy \(logic\) question by \(\pi\)... I'd rather notLog in to reply

– Sharky Kesa · 1 year, 4 months ago

Why? Those questions are awesome. Plus, this question shouldn't be too difficult.Log in to reply

– Sharky Kesa · 1 year, 11 months ago

Ahh... but that was only the first part o the problem. You have to find as many different solutions as you can.Log in to reply

– Jeremy Bansil · 1 year, 11 months ago

Nice!...Log in to reply

– Julian Poon · 1 year, 11 months ago

You should try it yourself! Its fun! :DLog in to reply