Salina wrote:

Hi Brent,

Is there any other way to solve this? I could not understand the process.

Thanks.

How about this?

9! = 9 x 8 x 7 x 6 x 5 x 4 x 2 x 1

= (3 x 3) x (2 x 2 x 2) x 7 x (2 x 3) x 5 x (2 x 2) x 2 x 1

= (2⁷)(3³)(5)(7)Once we know the prime factorization we can see that:

1, 2, 3, 4, 5, 6, 7, 8, and 9 are all factors of 9!

Also, 10 is a factor of 9! since 10 = (5)(2), and we can see one 5 and one 2 hiding in the

prime factorization of 9!Also, 12 is a factor of 9! since 12 = (2)(2)(3), and we can see two 2's and one 3 hiding in the

prime factorization of 9!14 is a factor of 9! since 12 = (2)(7), and we can see one 2's and one 7 hiding in the

prime factorization of 9!16 is a factor of 9! since 16 = (2)(2)(2)(2), and we can see four 2's hiding in the

prime factorization of 9!18 is a factor of 9! since 18 = (2)(3)(3), and we can see one 2 and two 3's hiding in the

prime factorization of 9!20 is a factor of 9! since 12 = (2)(2)(5), and we can see two 2's and one 5 hiding in the p

prime factorization of 9!22 is NOT a factor of 9! since 22 = (2)(11), and we there are NO 11's hiding in the

prime factorization of 9!Does that help?

Cheers,

Brent

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Brent Hanneson – Creator of greenlighttestprep.com

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