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How can I get the smallest positive number $p$ such that $10500p$ is a perfect cube?

I have tried to express $10500$ into its prime factors but not getting correct solution. Please help

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  • $\begingroup$ What solution do you get and how? $\endgroup$
    – Martin R
    Commented Mar 24, 2018 at 12:19
  • $\begingroup$ Well, $p = 1/10500$ is pretty small and positive, but $p = 1/(10500)^4$ is smaller. $\endgroup$
    – B. Goddard
    Commented Mar 24, 2018 at 12:27

2 Answers 2

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Calling $n=10500$ we need $m$ such that $$n p = m^3$$

now considering the prime factorization $m = \Pi_{k=1}^{\Phi} a^{\alpha_k}$

then $m^3 = \Pi_{k=1}^{\Phi} a^{3\alpha_k} = \Pi_{j=1}^{\Psi}b^{\beta_j} = n p$

Here $\Pi_{j=1}^{\Psi}b^{\beta_j}$ is the prime factorization for $n p$

but here $\Pi_{k=1}^{\Phi} a^{3\alpha_k} =\Pi_{j=1}^{\Psi}b^{\beta_j} = 2^2 \times 3 \times 5^3 \times 7 \times p$

hence $p = 2 \times 3^2 \times 7^2 = 882$

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    $\begingroup$ Thank you for contributing this correct answer and helping the OP. I've upvoted. But If you compare your work with the other (accepted) correct answer you will see that his is much easier to read and understand. Your's is right, but you use so much notation that it's hard for a beginner to follow. $\endgroup$
    – Ethan Bolker
    Commented Mar 24, 2018 at 12:49
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Since $10500 = 2^2 \times 3 \times 5^3 \times 7$, to make $10500p$ a perfect cube, the smallest possible $p$ is $2 \times 3^2 \times 7^2 = 882$, so that $$10500p = (2 \times 3 \times 5 \times 7)^3 = 210^3.$$

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