How many digits of 0 does 500! end with?

Question

Well the problem have to do with factorials

The expression n! is read "n factorial": and means n(n-1)(n-2)(n-3)...Thus, 6! meanse 6*5*4*3*2*1=720 and 10! means 10*9*8*7*6*5*4*3*2*1=3,628,800.

Notice that 6! ends with one digit of zero and 10! ends with 2 digits of 0. How many digits of 0 does 500! end with.

plz help ><

Answer 1

You get a zero for every factor of 5 multiplied by a factor of 2. So count the number factors of 5 in 500!, and the number of factors of 2 in 500!. The smaller of those will be the number of zeros at the end of 500!.

The smaller will be the number of factors of 5. We start with all multiples of five: they are 5, 10, 15, 20, 25, .... , 500.

There are one hundred of these, so there are 100 values with at least 1 factor of 5.

1 / 5 of these will have a second factor of 5; thats 20 more

1 /5 of THOSE will contain a third factor of 5; that's 4 more.

100 + 20 + 4 = 124 factors of 5 and thus 124 trailing zeros.

Answer 2

500! ends with 124 zeros