Bunuel wrote:
What is the greatest value of x such that 8^x is a factor of 18! ?
A. 1
B. 2
C. 4
D. 5
E. 6
Since 8 = 2^3, we are actually trying to determine the greatest value of x such that 2^(3x) is a factor of 18!.
Let’s first determine the number of factors of 2 within 18!. To do that, we can use the following shortcut in which we divide 18 by 2, and then divide the quotient of 18/2 by 2 and continue this process until we can no longer get a nonzero integer as the quotient.
18/2 = 9
9/2 = 4 (we can ignore the remainder)
4/2 = 2
2/2 = 1
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 18!.
Thus, there are 9 + 4 + 2 + 1 = 16 factors of 2 within 18!
However, we are not asked for the number of factors of 2; instead we are asked for the number of factors of 8. We see that 16 factors of 2 will produce 5 factors of 8.
Note: To clarify the final answer, note that the 16 factors of 2 can be expressed as 2^16. We now must break this number 2^16 into as many factors of 8 as possible; thus, we will have
2^16 = 2^3 x 2^3 x 2^3 x 2^3 x 2^3 x 2^1
2^16 = 8 x 8 x 8 x 8 x 8 x 2
2^16 = 8^5 x 2
Note that we can get only 5 factors of 8 out of 2^16; there is a “leftover” 2 that cannot be used.
Answer: D
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