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# Number Theory

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In the prime factorization of 12! + 15! + 18! + 21! + 24!, what is the exponent of 3?

Feb 27, 2022

#1
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If we divide 12! from both each term, 12!/12! = 1, 15!/12! will have a factor of 3, 18!/12! will have a factor of 3, 21!/12! will have a factor of 3, and 24!/12! will also have a factor of 3.

So thie expression will turn into

12!(1 + some multiple of 3 + some multiple of 3 + some multiple of 3 + some multiple of 3)

1 + some multiple of 3 + some multiple of 3 + some multiple of 3 + some multiple of 3 = not a multiple of 3

So we have 12! * not a multiple of 3.

Therefore, the exponent of 3 of 12! + 15! + 18! + 21! + 24! is the same as the exponent of 3 for 12!.

12! = 1*2*3*4*5*6*7*8*9*10*11*12, (3, 6, 9, 12) are our numbers divisable by 3.

3 = 3

6 = 2*3

9 = 2*3^2

12 = 2^2*3

Therefore, 1 + 1 + 2 + 1 = 5 is the exponent of 3.

I'm sorry if the explanation is confusing, please ask if there are any questions.

=^._.^=

Feb 27, 2022
#2
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12! + 15! + 18! + 21! + 24! ==

6.204994990790930078976e+23 = 2^10 * 3^5 * 5^2 * 7 * 11 * 53 * 129169 * 189221423

Feb 27, 2022