What is the greatest value of x such that 8^x is a factor of 18! ?

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