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 \(67108864\).

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 \(67108864\).

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## Comments

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TopNewestWe 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 · 2 years, 2 months ago

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Guts to powers – Akram Hossain · 2 years ago

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Sorry, note edited. – Bryan Lee Shi Yang · 2 years, 2 months ago

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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 · 2 years, 2 months ago

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

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