Select the correct answerHow many values can natural number n take, if n! is a multiple of 220 but not 320?radio_button_unchecked23radio_button_unchecked21radio_button_unchecked30radio_button_unchecked42

To solve this problem, we need to factorize 220 and 320.

220 = 2^2 * 5 * 11 320 = 2^6 * 5

For n! to be a multiple of 220, n must be at least 11 (to include the factor of 11).

For n! to not be a multiple of 320, n must be less than 16 (because 2^6 requires at least 16).

So the possible values for n are 11, 12, 13, 14, and 15.

Therefore, the correct answer is 5. However, this option is not given in the choices. There might be a mistake in the question or the options provided.

To solve this problem, we need to understand the prime factorization of 220 and 320.

The prime factorization of 220 is 2^2 * 5 * 11 and for 320 it is 2^6 * 5.

For n! to be a multiple of 220, n must be at least 11 (since 11 is the highest prime factor of 220).

However, for n! to not be a multiple of 320, n must be less than 16 (since 2^6 is part of the prime factorization of 320 and 2^6 is first achieved at 16!).

Therefore, the possible values for n are 11, 12, 13, 14, and 15.

So, the correct answer is none of the given options.