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(A) We are looking for natural numbers $n$ such that $n! + 10 = k^2$ for some integer $k$.Case $1$: If $n=1$,$1! + 10 = 11$ (not a perfect square).Cas

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(A) We are looking for natural numbers $n$ such that $n! + 10 = k^2$ for some integer $k$.Case $1$: If $n=1$,$1! + 10 = 11$ (not a perfect square).Case $2$: If $n=2$,$2! + 10 = 12$ (not a perfect square).Case $3$: If $n=3$,$3! + 10 = 6 + 10 = 16 = 4^2$ (a perfect square).Case $4$: If $n=4$,$4! + 10 = 24 + 10 = 34$ (not a perfect square).Case $5$: If $n=5$,$5! + 10 = 120 + 10 = 130$ (not a perfect square).Case $6$: If $n \ge 5$,then $n!$ is a multiple of $10$ because $n!$ contains the factors $2$ and $5$.Thus,$n! + 10 = 10m + 10 = 10(m+1)$ for some integer $m$.For $n \ge 5$,$n!$ is a multiple of $100$ (since $n!$ contains $2 	imes 5 	imes 10 = 100$ for $n \ge 10$,and for $n=5, 6, 7, 8, 9$,we check manually: $5!=120, 6!=720, 7!=5040, 8!=40320, 9!=362880$).Specifically,for $n \ge 5$,$n! + 10$ ends in the digit $0$ (since $n!$ ends in $0$ for $n \ge 5$).For a number ending in $0$ to be a perfect square,it must end in $00$. However,$n! + 10$ for $n \ge 5$ ends in $10, 20, 30, 40, 50, 60, 70, 80, 90$ (specifically $130, 730, 5050, 40330, 362890$).None of these are perfect squares.Therefore,only $n=3$ satisfies the condition.

How many natural numbers $n$ are there such that $n! + 10$ is a perfect square?

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