NMTC Inter Level Problem 7

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

Options:

(A) $0$

(B) $1$

(D) $5$

#Algebra

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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`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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?

mathh mathh - 6 years, 9 months ago

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

Nanayaranaraknas Vahdam - 6 years, 9 months ago

well, now you know.

mathh mathh - 6 years, 9 months ago

@Mathh Mathh – Could you also solve the other NMTC problem, which is a note?

Nanayaranaraknas Vahdam - 6 years, 9 months ago

@Nanayaranaraknas Vahdam – Aditya Raut has already posted a correct solution.

mathh mathh - 6 years, 9 months ago

I could do think of 2

HanAin AgHa - 6 years, 2 months 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}`	$\boxed{123}$