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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