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# NMTC Inter Level Problem 7

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

Options:

(A) $$0$$

(B) $$1$$

(C) $$2$$

(D) $$5$$

Note by Nanayaranaraknas Vahdam
2 years, 9 months ago

## Comments

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All 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? · 2 years, 9 months ago

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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. · 2 years, 9 months ago

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well, now you know. · 2 years, 9 months ago

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Could you also solve the other NMTC problem, which is a note? · 2 years, 9 months ago

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Aditya Raut has already posted a correct solution. · 2 years, 9 months ago

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I could do think of 2 · 2 years, 1 month ago

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