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Find the smallest integer $k$ such that $\exists n\in\mathbb Z$ with $2^{10}\cdot3^4\cdot5\cdot k = n^6$.

I have no sweet clue how to do this question.

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Well, let $k=0$. –  André Nicolas Dec 16 '12 at 23:31
    
I was just going to find the smallest positive integer using FTA until I saw your comment –  Amr Dec 16 '12 at 23:32
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3 Answers

up vote 2 down vote accepted

Given any $n\in\Bbb Z$, we have $n^6\ge 0$, yes? Observe that this rules out any negative $k\in\Bbb Z$ as a possibility. Putting $k,n=0$ gives you the smallest case, then.

Now, if you're looking for a $k>0$, then we'll proceed another way. Note that if there is any prime $p$ such that $p\mid n^6$, then $p\mid n$, too. In particular, each of $2,3,5\mid n$, and to minimize $n>0$ (thereby minimizing $k>0$), we won't want any other prime factors.

Since we're only allowing $2,3,5$ for prime factors of $n$, and since $2,3,5$ all must divide $n$, then $$n=2^{m_1}\cdot 3^{m_2}\cdot 5^{m_3}$$ for some positive integers $m_1,m_2,m_3$. The prime factorization of $n^6$, then, will be of form $$n^6=2^{6m_1}\cdot 3^{6m_2}\cdot 5^{6m_3}.$$ Since $$2^{10}\cdot 3^4\cdot 5^1\cdot k=2^{6m_1}\cdot 3^{6m_2}\cdot 5^{6m_3}$$ implies $$k=2^{6m_1-10}\cdot 3^{6m_2-4}\cdot 5^{6m_3-1}=2^{6(m_1-2)+2}\cdot 3^{6(m_2-1)+2}\cdot 5^{6(m_3-1)+5},$$ and since the exponents can't be negative (if $k$ is to be an integer), then to make $k$ as small as possible, we'll need $m_1=2$ and $m_2,m_3=1$, whence it follows that $$k=2^2\cdot 3^2\cdot 5^5.$$

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Thanks, the n^6 = 2^6m... equation is confusing me ever so slightly. –  Doug Smith Dec 16 '12 at 23:48
    
Gotcha. I'll flesh that out. –  Cameron Buie Dec 16 '12 at 23:54
    
Thank you, perfect. –  Doug Smith Dec 17 '12 at 0:11
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Hint: Note that the exponent of every prime $p$ in $n^6$ is a multiple of $6$. What does this imply for $p=2,3,5$? What is the best choice of exponent of $p$ in $k$ for $p=2,3,5$ and what for $p\ge7$?

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What does "the exponent on every prime p" mean? –  Doug Smith Dec 16 '12 at 23:36
    
If $m=p^r\cdot s$ with $s$ not divisible by $p$, then I call $r$ the exponent of $p$ in $m$. –  Hagen von Eitzen Dec 16 '12 at 23:38
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Let $k = 2^{a} 3 ^b 5 ^c m$ where $(m,2) = (m,3) = (m,5) = 1$. Note that since we want the minimum $k$, we have $m=1$, then $$n^6 = 2^{a+10} 3^{b+4} 5^{c+1}$$ Hence, $$n = 2^{(a+10)/6} 3^{(b+4)/6} 5^{(c+1)/6}$$ This means $a,b \equiv 2 \pmod 6$ and $c \equiv 5 \pmod6$ and $a,b,c \geq 0$. Since we want the minimum $k$, we get $a=2$, $b=2$ and $c=5$. Hence, $$k = 2^2 3^2 5^5$$

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