Sum of Factorials

Question

What is the hundreds digit when 2014! + 2013! + ... + 3! + 2! + 1! is expressed as an integer?

I was hoping to find out some sort of pattern by trying the first few factorials, but far as I can see there's none. Also, there seems to be some formula for this, but I'm not really sure if I understand them. How would I solve this?

Answer 1

Notice that for $n\ge15$ the factorial will end upto $000$ ,

($15!$ or more contain $125=5^3$ as a factor and when this multiplied by $2^3$ will lead to $000$ at the end)

so we need to find hundreds digits of $14!+13!+12!+......+1!$ and this is equal to $313$ because the sum is $93928268313$ and so the hundreds digits of $$2014! + 2013! + ... + 3! + 2! + 1!$$is 313.

Answer 2

You can find the hundreds digit of a number if you know the remainder when divided by $1000$. Can you see that if $k$ divides $n_0!$ then $k$ divides $n! \, \, \forall n \geq n_0$?

If that's clear then note that $15!$ is divisible by $1000$ and you have your answer.

Answer 3

Hint:

Compute $15!$ and note the three last digits.

Answer 4

Perform the computations modulo $1000$ (no calculator required):

$$\begin{align}1!&\equiv1,\\ 2!&\equiv2,\\ 3!&\equiv6,\\ 4!&\equiv24,\\ 5!&\equiv120,\\ 6!&\equiv720,\\ 7!&\equiv40,\\ 8!&\equiv320,\\ 9!&\equiv880,\\ 10!&\equiv800,\\ 11!&\equiv800,\\ 12!&\equiv600,\\ 13!&\equiv400,\\ 14!&\equiv600,\\ 15!&\equiv0. \end{align}$$

Then the sum (easier backward)

$$313.$$