A simple question ..

The zeros at the end will be due to 5s and 2s in 1000! . Number of 2s in 1000! Is more than 500. (1000!/2) But, the number of 5s are lesser . So the limiting agent is 5. Calculating the numbers of 5s will give us the number of zeros at the end of 1000! Lets find out number of 5s in 1000! Which can be calculated quite easily: 1000/5=200 200/5=40 40/5=8 8/5=1. Total=200+40+8+1=249. This can be seen as similar to remainder theorem. Difference being: we are calculating quotients. We may call it quotient theorem.

Hence, the number of zeros at the end of 1000! Would be 249.

Okay, there are 1000 ÷ 5 = 200 multiples of 5 between 1 and 1000. The next power of 5, namely 25, has 1000 ÷ 25 = 40 multiples between 1 and 1000. The next power of 5, namely 125, will also fit in the expansion, and has 1000 ÷ 125 = 8 multiples between 1 and 1000. The next power of 5, namely 625, also fits in the expansion, and has 1000 ÷ 625 = 1.6... um, okay, 625 has 1 multiple between 1 and 1000. (I don't care about the 0.6 "multiples", only the one full multiple, so I truncate my division down to a whole number.)

In total, I have 200 + 40 + 8 + 1 = 249 copies of the factor 5 in the expansion, and thus:

    249 trailing zeroes in the expansion of 1000!

We have: 1000! have x zeros in the end, therefore 5^x is divisible by 1000! We see: From 1 to 1000, there are 200 numbers which leave the remainder of 0 when divided by 5^1; there are 40 numbers which leave the remainder of 0 when divided by 5^2; there are 8 numbers which leave the remainder of 0 when divided by 5^3 and there is 1 number which leave the remainder of 0 when divided by 5^4. Hence, x=200+40+8+1=249. The answer is 249.

Every 5! gives an extra zero, so use the greatest integer function to determine how many of each powers of 5 could go into 1000.