Bunuel wrote:

22! is divisible by x but not by x^2. If x is a prime number, what is the range of possible values of x?

A. 4

B. 6

C. 8

D. 16

E. 17

First, we can list all of the prime numbers below 22, to make it divisible by X.

- We get a set {2,3,5,7,11,13,17,19}

Second, we must evaluate whether \(X^2\) is hiding in the 22!.

- 2 : \(X^2\) hiding all of the even numbers in the 22!

- 3 : \(X^2\) --> we have more than 2 multiple of 3 in 22!

- 5 : \(X^2\) --> we have more than 2 multiple of 5 in 22!

- 7 : \(X^2\) --> we have more than 2 multiple of 7 in 22!

- 11 : \(X^2\) --> we have exactly 2 multiple of 11 in 22!

- 13 : \(X^2\) --> we have only 1 multiple of 13 in 22!, hence 22! is not divisible by \(13^2\)

- 17 : \(X^2\) --> we have only 1 multiple of 17 in 22!, hence 22! is not divisible by \(17^2\)

- 19 : \(X^2\) --> we have only 1 multiple of 19 in 22!, hence 22! is not divisible by \(19^2\)

Difference : 19-13 = 6.

B.

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