# Simple number puzzle

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.

Note by Sharky Kesa
5 years, 10 months ago

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Answer 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?

- 5 years, 10 months ago

Nice!...

- 5 years, 10 months ago

You should try it yourself! Its fun! :D

- 5 years, 10 months ago

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

- 5 years, 10 months ago

Try this

$9$_ $9$ _ $9 = 10$

- 5 years, 3 months ago

$-(-9\div9-9)$

- 5 years, 3 months ago

Sir, you are genius!

- 5 years, 3 months ago

But $--$ is $+$...

- 5 years, 3 months ago

Try the extension.

- 5 years, 3 months ago

After the crazy $logic$ question by $\pi$... I'd rather not

- 5 years, 3 months ago

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

- 5 years, 3 months ago