The number of natural numbers \('n'\) for which \(n!+5\) is a perfect cube is

**Options:**

(A) \(0\)

(B) \(1\)

(C) \(2\)

(D) \(5\)

The number of natural numbers \('n'\) for which \(n!+5\) is a perfect cube is

**Options:**

(A) \(0\)

(B) \(1\)

(C) \(2\)

(D) \(5\)

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TopNewestAll the cubic residues \(\pmod 9\) are \(0,1,8\), but \(n!+5\equiv 5\pmod {9}, \forall n\ge 6\), hence \(n\le 5\). After checking all the possible cases, only \(n=5\) works (\(5!+5=125=5^3\)), hence the answer is \(\boxed{(\text{B})\text{ }1}\). BTW, why did you post this as a note instead of a problem for points?

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I got the answer as 1, but was not sure whether it was right. I can't make a problem with the wrong solution.

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well, now you know.

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I could do think of 2

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