m is a positive integer. m! is ending with k zeros, (m+2)!
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.
Answer/Solution
6
Steps/Work
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 D.
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 D.