It's Almost 2014!

$n! - (n-1)! + (n-2)! - (n-3)! + ... + 4! - 3! + 2! - 1! =$

$n * (n-1)! -(n-1)! + (n-2) * (n-3)! - (n-3)! +...+4*3! - 3! + 2*1! - 1! =$

$(n-1)*(n-1)! + (n-3)*(n-3)! +...+ 3*3! + 1*1!$

So the problem is reduced to: $2013 * 2013! + 2011 * 2011! +...+3 * 3!+1 * 1!$

Since we only need the last three digits we only have to calculate till 13! since 15! contains 3 zeros, further factorials won't change the last 3 digits. $13 * 13! + 11 * 11! + 9 * 9! + 7 * 7! + 5 * 5! + 3 * 3! + 1 * 1! = 6509177019$

The answer is $\boxed{19}$

We are only asked for the last three digits; note that if $n \ge 15,$ then $n!$ has at least three factors of $5$ and three factors of $2,$ so it ends in $3$ zeroes. Thus, we may disregard the term $n!$ when $n \ge 15.$ Our sum becomes

$14! -13!+12!-11!+10!-9!+8!-7!+6!-5!+4!-3!+2!-1!.$

Computing the last three digits of these first few factorials, we get the following list:

• $1! = 1$ ends in $001.$

• $2! = 2$ ends in $002.$

• $3! = 6$ ends in $006.$

• $4! = 24$ ends in $024.$

• $5! = 120$ ends in $120.$

• $6! = 720$ ends in $720.$

• $7!$ ends in $040.$

• $8!$ ends in $320.$

• $9!$ ends in $880.$

• $10!$ ends in $800.$

• $11!$ ends in $800.$

• $12!$ ends in $600.$

• $13!$ ends in $800.$

• $14!$ ends in $200.$

So, the last three digits of our sum are

$200-800+600-800+800-880+320-40+720-120+24-6+2-1$

which equals $\boxed{19}.$

Note that: $2014!-2013!+2012!-\cdot\cdot\cdot-3!+2!-1!=\sum_{n=1}^{2014} \left(-1\right)^n \cdot n!$

Since $15!$ is the first factorial number containing $3$ trailing zeroes, and hence $1000\mid 15!$ , we get the following: $\sum_{n=1}^{2014} \left(-1\right)^n \cdot n! \equiv \sum_{n=1}^{14} \left(-1\right)^n \cdot n!\pmod{1000}$

The rest is a bit calculator work as I couldn't find a way to calculate the value more easily or manually: $\sum_{n=1}^{14} \left(-1\right)^n \cdot n!=81393657019 \equiv \fbox{019} \pmod{1000}$

We only care about the last three digits of the sum, which are the hundreds digit, tens digit, and ones digit. We also know that anything above $15! = 1307674368000$ ends in three zeroes (since it contains three factors of $5$ from $5$ , $10$ , and $15$ and it also contains eleven factors of $2$ from all the even numbers in its factorization). Taking the factorial of any integer higher than $15$ is multiplying $15!$ by some other integer, therefore, we can ignore every element of the sequence except those less than 15. If we then evaluate the expression $14! - 13! + 12! - 11! + 10! - 9! + 8! - 7! + 6! - 5! + 4! - 3! + 2! - 1!$ , we find an answer of $81393657019$ , which, if we take the last three digits is $\boxed{019}$ .

We can simply the problem above become 2013.2013! + 2011.2011! + ... + (2n+1).(2n+1)! + .... + 3.3! + 1.1!, Because three last digits 000 in 15.15!, so we only sum the three last digit of 13.13! + 11.11! + 9.9! + 7.7! + 5.5! + 3.3! + 1.1! = ..400 + ..800 + .. 920 + ..280 + 600 + 18 + 1 = ..3019. So, We have 019 is the three last digits. Answer : 19