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
3 years, 11 months ago

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1 vote

  Easy Math Editor

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Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
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\boxed{123} \( \boxed{123} \)

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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?

Julian Poon - 3 years, 11 months ago

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Try this

\( 9 \)_ \( 9\) _ \(9 = 10\)

Without addition.

Krishna Sharma - 3 years, 4 months ago

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

Julian Poon - 3 years, 4 months ago

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@Julian Poon Sir, you are genius!

Krishna Sharma - 3 years, 4 months ago

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@Krishna Sharma But \(--\) is \(+\)...

Julian Poon - 3 years, 4 months ago

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@Julian Poon Try the extension.

Sharky Kesa - 3 years, 4 months ago

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@Sharky Kesa After the crazy \(logic\) question by \(\pi\)... I'd rather not

Julian Poon - 3 years, 4 months ago

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

Sharky Kesa - 3 years, 4 months ago

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

Sharky Kesa - 3 years, 11 months ago

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Nice!...

Jeremy Bansil - 3 years, 11 months ago

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You should try it yourself! Its fun! :D

Julian Poon - 3 years, 11 months ago

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