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.