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# multiples

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

Jan 7, 2021

#1
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\$\$12/3 = 4, 12/9 ≈ 1.\$\$

There are 5 powers of 3 in \$12!\$

\$\$15/3=5, 15/9 ≈1.\$\$

There are 6 powers of 3 in \$15!\$

\$\$18/3 = 6, 18/9 =2.\$\$

There are 8 powers of 3 in \$18!\$

You can likewise figure out that there are 9 in \$21!\$ and 10 in \$24!.\$ Thus, as there are 5 numbers, there will be no multiplying of 3 within non-3-divisible numbers, so the answer will relate to the sum \$3^5 + 3^6 + 3^8 + 3^9 + 3^{10}.\$ This is just \$3^5(1 + 3 + 27 + 81 + 243,)\$ or \$3^5(355),\$ so the answer is \$\boxed{5}.\$

Jan 8, 2021
edited by Pangolin14  Jan 8, 2021