ardz24 wrote: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!/2
¹� is an integer, where 10 is greatest possible value for the exponent in blue.
12!/2¹� = 12!/(2²)� = 12!/4�.
Since 12!/2
¹� 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.
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