Huge easy difficulty.

Find minimum number $n$, such that

$2^{n!}$ × $2^{n-1!}$ × $2^{n-2!}$ ×... ...× $2^{3!}$ × $2^{2!}$ × $2^{1!}$

is a perfect power of $67108864$ .

#NumberTheory

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

We start by observing that $67108864 = 2^{26}$

So, it suffices to find the least $n$ such that $\sum_{r=1}^n r! = 26k$ for some arbitrary positive integer $k$ .

But observe that $\sum_{r=1}^n r! = 1+ \sum_{r=2}^n r! = 1+2N$

Hence, we see that the LHS is odd whereas RHS is even. And therefore, no solution exists for $n$ .

Kishlaya Jaiswal - 6 years, 3 months ago

I don't understand your notation. Could you be more explicit with your product formula?

D G - 6 years, 3 months ago

Assuming you mean to write parentheses around your exponents, there is no solution. Your question is equivalent to asking if there are any integers which satisfy 2^x = 9615^y, which is false since gcd(2, 9615) = 1.

D G - 6 years, 3 months ago

Sorry, note edited.

Bryan Lee Shi Yang - 6 years, 3 months ago

Guts to powers

Akram Hossain - 6 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$