SANSKAR'S OG PROBLEMS

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Problem1

Let $\overline{ab}$ be a 2-digit positive integer satisfying $\overline{ab}^2$ = $a! +b!$ . Find $a+b$ .

Solution 1 by ddk001

we have

$10a+b=a!+b!$

Obviously, the left increase much slower for big $a$ and $b$ so $a$ and $b$ must be small. (Namely, $a!+b!<100 \implies a,b<5$ ) $a=b=4$ don't work so $a!+b! \le 3!+4!=30$ , so since $a,b<5$ , the max of $a!+b!$ is $24$ , which didn't work so $a,b<4$ . since the max now would be $12=3!+3!$ , and that doesn't work, the max is now $2!+3!=8$ , which is not a 2-digit integer. Hence, no such ordered pair $(a,b)$ exist. ~Ddk001

Problem2

For any positive integer $n$ , $n$ >1 can $n!$ be a perfect square? If yes, give one such $n$ . If no, then prove it.

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SANSKAR'S OG PROBLEMS

Problem1

Solution 1 by ddk001

Problem2