2010 AMC 12B Problems/Problem 9

Problem

Let $n$ be the smallest positive integer such that $n$ is divisible by $20$, $n^2$ is a perfect cube, and $n^3$ is a perfect square. What is the number of digits of $n$?

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$

Solution

We know that $n^2 = k^3$ and $n^3 = m^2$. Cubing and squaring the equalities respectively gives $n^6 = k^9 = m^4$. Let $a = n^6$. Now we know $a$ must be a perfect $36$-th power because $lcm(9,4) = 36$, which means that $n$ must be a perfect $6$-th power. The smallest number whose sixth power is a multiple of $20$ is $10$, because the only prime factors of $20$ are $2$ and $5$, and $10 = 2 \times 5$. Therefore our is equal to number $10^6 = 1000000$, with $7$ digits $\Rightarrow \boxed {E}$.

See also

2010 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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