×

# 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 $$67108864$$.

Note by Bryan Lee Shi Yang
2 years, 4 months ago

Sort by:

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 $$n$$. · 2 years, 4 months ago

Guts to powers · 2 years, 2 months ago

Sorry, note edited. · 2 years, 4 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. · 2 years, 4 months ago

I don't understand your notation. Could you be more explicit with your product formula? · 2 years, 4 months ago

×