What is the smallest possible value of the positive integer \(n\) such that \(2016n\) is a perfect cube?

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# A Timely Cube

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Solution

Posted on by Mr Haines

What is the smallest possible value of the positive integer \(n\) such that \(2016n\) is a perfect cube?

Lots of good solutions for this one. Most of us noticed that in the prime factorisation of any cube number \(N\)

\[ N = \prod_{i=1}^{m} p_{i}^{k_{i}} \]

we must have each \(k_{i}\) a multiple of \(3\). Since \(2016 = 2^{5} \times 3^{2} \times 7\) we need to multiply by \(n = 2 \times 3 \times 7^{2} = 294\) in order to get the cube number

\[ N = 2^{6} \times 3^{3} \times 7^{2} = 74088\]

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