What is the greatest value y for which 4^y is a factor of 12

[BTGmoderatorAT]

What is the greatest value y for which 4^y is a factor of 12!?

A. 2
B. 3
C. 4
D. 5
E. 6

Is there a strategic approach to this question? Can any experts assist?

[GMATGuruNY]

What is the greatest value y for which 4^y is a factor of 12!?

A. 2
B. 3
C. 4
D. 5
E. 6

To determine the greatest possible value for y, we need to know how many 4's can divide into 12!.
The number of 4's that can divide into 12! depends on the NUMBER OF 2's contained within the prime-factorization of 12!.
To count the number of 2's contained within 12!, divide increasing POWERS OF 2 into 12.

Every multiple of 2 contained within 12! provides at least one 2:
12/2 = 6 --> 6 2's.
Every multiple of 2² provides a SECOND 2:
12/2² = 3 --> 3 more 2's.
Every multiple of 2³ provides a THIRD 2:
12/2³ = 1 --> 1 more 2.
Thus, the total number of 2's contained within 12! = 6+3+1 = 10.

Since there are a total of ten 2's contained within 12!, we get:
12!/2¹� is an integer, where 10 is greatest possible value for the exponent in blue.
12!/2¹� = 12!/(2²)� = 12!/4�.
Since 12!/2¹� is an integer (where 10 is greatest possible value for the exponent in blue), 12!/4� must also be an integer (where 5 is greatest possible value for the exponent in red).
Thus, the greatest possible value for y is 5.

The correct answer is D.

Since 12! = 1*2*3*4*5*6*7*8*9*10*11*12, an alternate way to count the number of 2's contained with 12! is to prime-factorize the even integers between 1 and 12, inclusive:
2 = 2
4 = 2*2
6 = 2*3
8 = 2*2*2
10 = 2*5
12 = 2*2*3.
The blue values above indicate that the number of 2's contained within 12! = 10.

[Scott@TargetTestPrep]

What is the greatest value y for which 4^y is a factor of 12!?

A. 2
B. 3
C. 4
D. 5
E. 6

Let's first determine the number of 2's in 12!.

To determine the number of 2s within 12!, we can use the following shortcut in which we divide 12 by 2, then divide the quotient of 12/2 by 2 and continue this process until we no longer get a nonzero quotient.

12/2 = 6

6/2 = 3

3/2 = 1 (we can ignore the remainder)

Since 1/2 does not produce a nonzero quotient, we can stop.

The next step is to add up our quotients; that sum represents the number of factors of 2 within 12!.

6 + 3 + 1 = 10

We see that there are ten 2's in 12!; however, since 4 = 2^2, and since 10/2 = 5, we see that there are five 4's in 12!.

Answer: D