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# 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
2 years, 11 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?

- 2 years, 11 months ago

Try this

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

- 2 years, 4 months ago

$$-(-9\div9-9)$$

- 2 years, 4 months ago

Sir, you are genius!

- 2 years, 4 months ago

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

- 2 years, 4 months ago

Try the extension.

- 2 years, 4 months ago

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

- 2 years, 4 months ago

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

- 2 years, 4 months ago

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

- 2 years, 11 months ago

Nice!...

- 2 years, 11 months ago

You should try it yourself! Its fun! :D

- 2 years, 11 months ago