# if 6^k and 15k are factors of 43!, then which of the following could be the values of k ?

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• 6^k = 2^k * 3^k

43! contains 14 multiples of 3, 4 multiples of 3^2 and, 1 multiple of 3^3. That's a total of 19 threes in the prime factorization of 42!.

There are even more twos in the prime factorization of 42! so the number of factors of 6 in 43! is 19. The most k can be is 19

15k = 3*5*k will be a factor of 43! for any number less than 20 since 3, 5 and k are all factors of 43! except when k is 3 or 5. But there will be two threes and two fives in the factors of 43! so that's not a problem either.

k can be anything from 1 to 19 inclusive.

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• 43! is calculated as 43 X 42 X 41.....X 3 X 2 X 1, and so contains all positive whole number integers including 43 as its factors.

6^k would equal 1, 6, 36, and 216, for k = 0, 1, 2 and 3, respectively. Of these, 1, 6 and 36 are factors of 43!, but k cannot be 3 because 216 is not one of the factors. So, k can only be 0, 1 and 2.

However, since 15(0) = 0, but 0 cannot be a factor of 43!, so k can equal only 1 and 2.

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