shahul wrote:
What is the maximum value of m such that 7^m divides into 14! evenly?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Note: 7^m divides 14!. Thus we understand that 7^m is basically a factor of 14!.
The you can find out the maximum number of m as a power of 7 as follows:
n/x^1 + n/x^2+......so on...
14/7
=2
m must be 2 as a power of 7.
Hope it clears.
Alternate way:
14!=1*2*3*34*5*6*7*8*9*10*11*12*13*13*14
=1*2*3*4*5*6*7*8*9*10*11*12*13*7*2
= \(7^2\)*1*\(2^2\)*3*4*5*6*8*9*10*11*12*13
So in total we have 2..........7