Factorials!

Since the last digit of $r!$ , when $r \geq 5$ , is zero, it is clear that for any $x\geq 4$ $\bigg(\sum_{r=1}^{x} r! \bigg)\mod 10 = \bigg(\sum_{r=1}^{4} r!\bigg) \mod 10 = 3$ Since the last digit of a perfect square can never be $3$ , it is clear that $x \leq 3$ . By direct computation we see that, for $x=3$ , the sum is a perfect square.

A much simpler way to see it is to notice that $1! + 2! + 3!$ = 9. Now we see that if we add $4!$ to 9, we get 33. Notice that $5!$ , $6!$ , $7!$ and so on always end in 0. That means the units digit will always stay at 3. No square numbers end in the digit 3, therefore making the answer 3.

This problem is not rated dificult but took a while for me to convince myself that I was not omitting the largest perfect square. The correct answer is x = 3.

This problem is not rated dificult but took a while for me to convince myself that I was not omitting the largest perfect square. The correct answer is x = 3.