m is a positive integer. m! is ending with k zeros, (m+2)! is ending with k+2 zeros. Find the number of possible values of m, if 90 ≤ m ≤ 190.  Choose the correct choice from the following choices
A) 2
B) 3
C) 4
D) 6
E) 8
Possible values of m where (m+2)! has two zeroes more than the no. of zeroes in m! are   m=98,99--No of zeroes 22 (m+2)! i.e, 100 and 101 have 24 zeroes each.   m=148,149   m=174,175   no. of possible values of m=6.   Answer has to be 6.
<result>D</result>
m is a positive integer. m! is ending with k zeros, (m+2)! is ending with k+2 zeros. Find the number of possible values of m, if 90 ≤ m ≤ 190.