Pete Inglesby from Wolverhampton; Patrick Snow, age 16, Dame Alice Owens School, Potters Bar; Yatir Halevi, age 18, Maccabim-Reut High-School, Israel; Andrei Lazanu, age 12, School 205, Bucharest, Romania all solved this problem. Here is Pete's solution:

Consider numbers of the form

$$ u(n) = 1! + 2! + 3! +...+ n!. $$

How many such numbers are perfect squares?

In fact only when $n=1$ or $n=3$ does $u(n)$ give a perfect square (1 and 9 respectively).

The solution lies in the fact that all factorials above $5!$ are multiples of 10. This is because they all have 2 and 5 as prime factors. For $n> 5$

\begin{matrix} n! & = 1 \times 2 \times 3 \times 4 \times 5 \times \dots \times n \\ = 2 \times 5 (1 \times 3 \times 4 \times \dots \times n) \\ = 10 (1 \times 3 \times 4 \times \dots n) . \end{matrix}

When adding a multiple of 10, to another integer, $k$ , the last digit of $k$ will remain the same. Suppose $k = 10 j + l$ and $m = 10 n$ , with $l$ and $n$ integers, then $k + m = 10 (j + n) + l$ .

So when adding all factorials equal to or higher than 5! then you are adding a multiple of ten, so the last digit of the sum will be the same. This is important because the last digit of $u (4)$ is a 3. We have $u (4) = 1! + 2! + 3! + 4! = 1 + 2 + 6 + 26 = 33$ . Therefore for any $u (n)$ with $n > 5$ the last digit will be 3. For any integer $p = (10 a + b)$ with $a$ and $b$ integers and $b < 10$ , $p^{2} = (100 a^{2} + 20 ab + b^{2})$ . Therefore the last digit is of $p^{2}$ is only dependent on $b^{2}$ .

b b ² Last digit of b ²

0 0 0

1 1 1

2 4 4

3 9 9

4 16 6

5 25 5

6 36 6

7 49 9

8 64 4

9 81 1

This table shows that the squares of all integers between 0 and 9, and therefore all integers, end in either 1, 4, 5, 6 or 9.

No perfect squares end in 3. Therefore, u(n) with n > 4 can never be a perfect square as it has been shown to always end in 3.

All that remains is to find u(n) for all integers of 4 or less:

u (1) = 1! = 1

;

$u (2) = 1! + 2! = 3$ ;

$u (3) = 1! + 2! + 3! = 9$ ;

$u (4) = 1! + 2! + 3! + 4! = 33 .$

Only u(1) and u(3) are perfect squares, and so they are the only sums of factorials to be perfect squares.

b	b ²	Last digit of b ²
0	0	0
1	1	1
2	4	4
3	9	9
4	16	6
5	25	5
6	36	6
7	49	9
8	64	4
9	81	1