# Thread: Factorials & Perfect Squares

1. ## Factorials & Perfect Squares

I've been racking my brain out for the last two days on this problem, but I don't even have the slightest clue on how to approach this problem. Could anybody point me in the right direction please?

Q: Find the values of $\displaystyle n \geq 1$ for which $\displaystyle 1! + 2! + ... + n!$ is a perfect square. (Hint: Determine the possibilities for the units digit of a perfect square.)

Using the hint:
$\displaystyle 0^2 = 0$
$\displaystyle 1^2 = 1$
$\displaystyle 2^2 = 4$
$\displaystyle 3^2 = 9$
$\displaystyle 4^2 = 16$
$\displaystyle 5^2 = 25$
$\displaystyle 6^2 = 36$
$\displaystyle 7^2 = 49$
$\displaystyle 8^2 = 64$
$\displaystyle 9^2 = 81$
$\displaystyle 10^2 = 100$

This leads to a pattern: 0 1 4 9 6 5 6 9 4 1 0

Then I'm just lost....

2. The unit digit of $\displaystyle 1!$ is 1.

The unit digit of $\displaystyle 2!$ is 2.

The unit digit of $\displaystyle 3!$ is 6.

The unit digit of $\displaystyle 4!$ is 4.

The unit digit of $\displaystyle n!, \ n\geq 5$ is 0.

Now, we have:
$\displaystyle 1!=1=1^2$

$\displaystyle 1!+2!+3!=1+2+6=9=3^2$

For $\displaystyle n\geq 4$ the unit digit of $\displaystyle 1!+2!+3!+\ldots+n!$ is 3 and it can't be a perfect square.

3. Thanks a lot red_dog.

After your hint, I completed the mathematical proof and finished off the problem.