Note that $n$ can also be expressed as $\sum_{x=1}^{100} x!$

Because the question asks for the last there digits, we will only focus on the last three digits.

Every fifth factorial gains a zero at the end. For example, $5! = 120$ has one zero at the end. $10! = 3,628,800$ has two zeros at the end. This pattern continues at every fifth factorial. Anything after $15!$ will have three zeros at the end and will not affect the sum.

Thus the last three digits of $\sum_{x=1}^{100} x!$ is equal to the last three digits of $\sum_{x=1}^{14} x!$

Therefore, you only need to find the sum of $1! + 2! + 3! + ... + 14!$ .

Adding up $1! + 2! + 3! + ... + 14!$ , you get a number ending in $313$ .

The answer is $\boxed{313}.$

Let's count the last three digits of the first few values
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7! = ...040
8! = ...320
9! = ...880
10! = ...800
11! = ...800
12! = ...600
13! = ...800
14! = ...200
15! = ...000

Since 15! to 100! have ...000 in their last three digits, then, we only need to sum up the last three digits of 1! until 14!

$n = 1! + 2! + 3! +... + 14! + ...$
$n = 1+2+6+24+120+720+...040+...320+...880+...800+...800+...600+...800+...200+ ...$
$n = ...313$

Hence, the last three digits of n is $\boxed{313}$

After 14! the last three digits are 0's. Hence it is enough to find the last three digits of 1!+2!...14! which is 313.

1!=1+ 2!=2+ 3!=6+ 4!=24+ 5!=120+ 6!=720+ 7!=5040+ 8!=40320+ 9!=362880 + 10!= 3628800 + 11!=39916800+ 12!=479001600+ 13!=6227020800 + 14!=87178291200+ 15!=1307674368000+

16!=...........................................(above 14! the last three digit will be 000)

answer= sum of last three digit of 1! to 14!=5313 n= last four digit is 5313 last three digit is 313

First note that the number of zeros at the end of $n!$ is the same as the amount of fives in its prime factorization. Since we know that the first factorial number to have three zeros is $15!$ we can simply take $1!+2!+3!+\cdots+14!$ mod 100. After a bit of bashing (or calculator work) you will find that the answer is 313.

Just find 1!+2!....+14! Because 15! Will have 000 at the end(2.5.10.(5).(2)) The last bracketted 5 and 2 are from 15 and any even no resp... simple...

1!=1+ 2!=2+ 3!=6+ 4!=24+ 5!=120+ 6!=720+ 7!=5040+ 8!=40320+ 9!=362880 + 10!= 3628800 + 11!=39916800+ 12!=479001600+ 13!=6227020800 + 14!=87178291200+ 15!=1307674368000+ 16!=...........................................(above 14! the last three digit will be 000) ================ answer= sum of last three digit of 1! to 14!=5313 n= last four digit is 5313 last three digit is 313

We know if 3- zero digits of the k! is 15!, so the last three digits of the 1! + 2! + ... + 100! = the last three digits of the 1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! + 10! + 11! + 12! + 13! + 14! =1 + 2 + 6 + 24 + 120 +720 + 040 + 320 + 880 + 800 + 800 + 600 + 800 + 200 = ....(313). Answer ; 313

10 X 15 X 20=3000

That's means that from 20! and on all the number are divided by 1000 and not relevant

Even if n was 1! + 2! + ....+ 1000000! all we have to find is the summary of 1! + 2! + .... 19!

I found that using a computer :)

For the given expression $n = 1! + 2! + 3! + 4! .... + 100!$

when you evaluate the values of all factorials, you will find that the exponential value of 10 will be equal to 3 for $15!$

i.e for $15!$ is a factor of 1000. Hence, to find out the last three digits of $n$ we need to evaluate

$1! + 2! + 3! + 4! +5! + ..... + 14! = 93928268313$

Hence, last three digits of $n$ is $\boxed{313}$

• 5! =...0
• 10! = ...00
• 15! = ....000
• n! = ...000 (at least 3 times zero '0') (n>15) => The last three digits of n is the same of 1! + 2! + 3! + ... 13! + 14! = ...313

Look at 15! cong to 0 mod 1000. Then sum 15! + 14! + 13! + 12! + ... + 2! + 1! cong to 100! + 99! + ... + 3! +2!+1! mod 1000