Question

what is the smallest positive integers of k if 3168k is a perfect cube​​

Answer 1

Answer:

What is the smallest positive integer 'K' such that (2000) (2001) K is a perfect cube?

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Asked to answer by .

This question already has been answered, so my answer won't be much different from others'. Here goes:

2000∗2001∗k is a perfect cube.

OR 24∗3∗53∗23∗29∗k is a perfect cube.

OR (23∗53)∗2∗3∗23∗29∗k is a perfect cube.

After pairing up the factors, we come to the conclusion that a 2 , a 3 , a 23 and a 29 remain without being able to be paired into triplets (by triplets of a number x here we mean that we have x∗x∗x as a part of the overall factorization).

Hence, if we complete the such triplets of the remaining unfortunate primes, we will find that (23∗53)∗(2∗22)∗(3∗32)∗(23∗232)∗(29∗292) becomes a perfect cube (because then the number 2000∗2001∗k can be decomposed into prime factors with multiple-of- 3 occurrences).

Thus, it can be concluded that the smallest value of k for which 2000∗2001∗k becomes a perfect cube is

k=22∗32∗232∗292=16016004

And the number, in turn, becomes k∗2000∗2001=64096048008000