Bunuel wrote:

What is the greatest value of x such that 9^x is a factor of 9! ?

A. 1

B. 2

C. 3

D. 4

E. 5

We need to determine the greatest values of x such that 9^x (or 3^2x) divides into 9!. In other words, we need to determine the number of factors of 9 in 9!. Since 9 = 3^2, let’s first determine the total number of factors of 3 in 9!.

To do so, we can use the following shortcut in which we divide 9 by 3, then divide the quotient (ignore any nonzero remainder) by 3 and continue this process until we no longer get a nonzero quotient.

9/3 = 3

3/3 =1

Since 1/3 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 3 within 9!.

Thus, there are 3 + 1 = 4 factors of 3 within 9!, and therefore there are 2 factors of 9 in 9!.

Alternate solution:

Recall that 9! = 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1. We see that the first factor 9 is a factor of 9. We also see that 6 x 3 = 18 will contain another factor of 9. Thus there are only 2 factors of 9 in 9! since none of the other numbers contain or contribute a factor of 9 or 3. Thus, the largest value of x such that 9^x is a factor of 9! is 2.

Answer: B

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