AndyB wrote:Hi All,
This is one of the Veritas Prep question.
Could someone plese explain it in more simple terms.
If 6^y is a factor of (10!)^2, what is the greatest possible value of y?
(A) 2
(B) 4
(C) 6
(D) 8
(E) 10
6^y = k. (10!)^2
y will be maximum when there are maximum 6's in (10!)^2
(10!)^2 = (10*9*8*7.....2*1) ^2
get pairs of 3 and 2.
9 = 3*3. in this case it is squared, (3*3) ^2
8 = 2*2*2. (2*2*2) ^2
6^2 as is.
3^2
2^2.
No multiples of 3 remain now.
Combine them to form 6's.
3^2, 3^2, 2^2,2^2 .... from 9 and 8, leave unmatched 2's, which is 2^2 in this case from 8.
Number of 3's and 2's are = 2+2 = 4 ..................................... (1)
3^2 and 2^2 (last 3 and 2 after expanding 10 !)
Number of 3's and 2's i.e. 6's = 2 ............................................(2)
6^2 ... as is.
Number of 6's = 2.....................................................................(3)
Combine 1,2 and 3.
4+2+2 = 8.
Pick D.
Hope is the dream of a man awake
