What is the smallest positive integer n for which n^(2) is divisible by 18 and n^(3) is divisible by 640?

Question

What is the smallest positive integer n for which
`n^(2)` is divisible by 18 and
`n^(3)` is divisible by 640?

Answer 1

120

Explanation:

n^2 has a factor of 18, so factors of 3^2·2. Since n^2 is a perfect square, we know n must have a factor of 3·2 = 6.

n^3 has a factor of 640, so factors of 2^7·5. Since n^3 is a perfect cube, we know n must have a factor of 2^3·5 = 40.

The least common multiple of 6 and 40 is 120.

The smallest positive integer n is 120.

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Check

120^2/18 = 800

120^3/640 = 2700