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I want to find the smallest positive integer A in which $$10A$$ is a perfect square and $$6A$$ is a perfect cube


Thanks for the hint, I can see now I just needed $$2^5,3^2 , 5^3$$

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  • $\begingroup$ Did you try to write $\;A^{10}\;$ ? If you did I think you really need to take a peek at meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Timbuc May 1 '15 at 14:36
  • $\begingroup$ Do you mean smallest positive integer? $\endgroup$ – Gregory Grant May 1 '15 at 14:36
  • $\begingroup$ If you did mean to say $A^{10}$ and $A^6$, then every $A$ works... (as $A^{10}=(A^5)^2$ and $A^6 = (A^2)^3$). Please clarify what exactly you mean by $A10$ and $A6$. $\endgroup$ – JMoravitz May 1 '15 at 14:38
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    $\begingroup$ I wonder if he means the number formed by the concatenation of $10$ to the digits of A, which results in a perfect square, and likewise, the concatenation of 6 to the digits of A (resulting in a perfect cube.) $\endgroup$ – Jordan Glen May 1 '15 at 14:40
  • $\begingroup$ no matter the interpretation this question does not belong under number-theory. // In the new version, I assume $10A$ means $A+ A + \ldots + A$ with 10 copies? $\endgroup$ – Willie Wong May 1 '15 at 14:43
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Hint: if you factor a square into prime factors, say $n^2=p_1^{a_1}p_2^{a_2}\dots$ where the $p_i$ are primes, all the $a_i$ will be even. Similarly for a cube, the $a_i$ will be multiples of $3$. Here your primes are only $2,3,5$

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