Problem Solving

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

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Moved this post from "GMAT Strategy"

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.

We need to find how many 3's are in 10!

10/3+10/3^2 =3+1 [Take only the quotient]

Four 3's are available in 10!

-->There are 8 3's available in (10!)^2

Pick D

Sorry for the wrong post.

Hi Seelango,

What i understand is - Its about finding (3 x 2) pairs in the series 10x9x8x7x6x5x4x3x2x1.
Here it is 4.Since we have 10!^2 it would become 8.
Please correct me if i am wrong.

Exactly.The prime factors of 6 are 2 and 3.

Since there are many 2's in 10!, we need to find how many 3's are in 10! so that we can find 3*2 pairs.

There are 4 3's in 10!.So totally 8 3's in (10!)^2.

Hope this clarify!!!!

Thanks a lot Anand