Bunuel
What is the greatest value of integer n such that 4^n is a factor of 26! ?
A. 6
B. 9
C. 10
D. 11
E. 12
Since 4 = 2^2, we are actually trying to determine the largest value of n such that 2^(2n) is a factor of 26!.
Let’s first determine the number of factors of 2 within 26!. To do that, we can use the following shortcut in which we divide 26 by 2, and then divide the quotient of 26/2 by 2 and continue this process until we can no longer get a nonzero integer as the quotient.
26/2 = 13
13/2 = 6 (we can ignore the remainder)
6/2 = 3
3/2 = 1 (we can ignore the remainder)
Since 1/2 does not produce a nonzero quotient, we can stop.
The final step is to add up our quotients; that sum represents the number of factors of 2 within 26!.
Thus, there are 13 + 6 + 3 + 1 = 23 factors of 2 within 26!
However, we are not asked for the number of factors of 2; instead we are asked for the number of factors of 4. Since there are 23 factors of 2 within 26!, and since 23/2 = 11 remainder 1, there are 11 factors of 4 with a factor of 2 left over.
In other words, 2^23 = 4^11 x 2 (which equals 2^22 x 2^1 = 2^23).
Answer: D