mohnish104 wrote:

What is the greatest integer m for which the number 50!/10^m is an integer?

(A) 5

(B) 8

(C) 10

(D) 11

(E) 12

Notice that m is basically the power of 10. let's analyse the character of 10 with exponents.

10^1=10

10^2=100

10^3=1000

10^4=10000

so, one thing is clear that any exponent of 10 increases the no of zeros.

ONE important point to be noted that for each zero, one 5 and one 2 are responsible.

50!/10^m will be an integer. to do so we have to eliminate 10^m. 10^m could be 100 , 1000, 1000000 and so on. we will get just zeros in 10^m. As we don't know the value of m it will depend on the no of zero 50! has. it will tell us how many zero we will need to vanish 10^m. Thus we have to find out the no. of zero in 50!.

there is a short-cut to find out zero:

50/5 + 50/(5)^2 = 10+ 2 = 12 . it means will get 12 zero in total in 50!. So, m will be 12.

Now why did we divide 50 by 5 . we have already learnt that each pair of 5 and 2 creates a zero. In 50! we will have no. of 2 as its factor but 5 is limited. So, we found out the no. of 5 in 50!. ultimately, we got 12 fives. thus , 12 pairs of 5 and 2 yield 12 zero.

Thus, the correct answer is E.