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Find a positive integer such that half of it is square,a third of it is a cube,and a fifth of it is a fifth power.

epimorphic
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saad abdurahmman
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    What have you tried? It better have factors of $2,3,5$, for example. How many of each? – Ross Millikan Jan 17 '16 at 22:17
  • This was re-asked as http://math.stackexchange.com/questions/1872090/find-a-number-such-that-half-of-it-is-a-square-a-third-of-it-is-a-cube-and-a-fif – Mark S. Jul 27 '16 at 01:39

1 Answers1

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Let $n = 2p^{2} = 3q^{3} = 5r^{5},$ where each of $n,$ $p,$ $q,$ and $r$ are positive integers. From this, we know that $n$ is divisible by each of $2,$ $3,$ and $5,$ so we can write further that $n = k \cdot 2^{a}3^{b}5^{c}.$

From the first equation, we know that $a$ is odd, and $b$ and $c$ are both even. In addition, $b$ is one more than a multiple of $3,$ while $a$ and $c$ are both divisible by $3.$ Finally, $c$ is one more than a multiple of $5,$ while $a$ and $b$ are both divisible by $5.$

By inspection, the minimum possible value of $a$ is $15.$ Similarly, the minimum possible value of $b$ is $10.$ The minimum possible value of $c$ is $6.$ The minimum possible value of $k$ is simply $1.$

The minimum possible solution is $\boxed{2^{15}3^{10}5^{6}}.$

K. Jiang
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  • you could, having decided on exponents $a,b,c$ arrive at your values by applying the Chinese remainder Theorem for each. Not hard in this case. – Will Jagy Jan 17 '16 at 23:26
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    Yes that works, but the numbers are low in this case, so I just solved it using inspection. @Will Jagy – K. Jiang Jan 18 '16 at 00:45