# 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 $\Rightarrow \boxed {E}$.

## Solution 2 (Chinese Remainder Theorem)

Let $n = 2^2 \cdot 5 \cdot x$ for some integer $x$, then we know that \begin{align*} n^2 = 2^4 \cdot 5^2 \cdot x^2 = m_0^3\\ n^3 = 2^6 \cdot 5^3 \cdot x^3 = m_1^2 \end{align*} Since we have $2^4 \cdot 5^2 \cdot x^2$ a power of 3, that means, by the Fundamental Theorem of Arithmetic, the exponents of the primes must each be divisible by 3. This also means that there's no point in $p \mid x$ if $p \not \in \{2,3\}$, since that would just serve to increase $n$ and we want $n$ to be minimal.

This means we can write $x = 2^{\alpha} \cdot 5^{\beta}$ and thus we have $2^4 \cdot 5^2 \cdot 2^{2\alpha} \cdot 5^{2\beta} = 2^{4 + 2\alpha} \cdot 5^{2 + 2\beta} = m_0^3$, therefore we know \begin{align*} 4 + 2\alpha \equiv 0 \pmod{3}\\ 2 + 2\beta \equiv 0 \pmod{3} \end{align*} and doing the same with the third equation being a square, we have \begin{align*} 6 + 3\alpha \equiv 0 \pmod{2}\\ 3 + 3\beta \equiv 0 \pmod{2} \end{align*} We can shift this around into two systems involving the same variables, as follows: \begin{align*} \alpha \equiv 0 \pmod{2}\\ 1 + 2\alpha \equiv 0 \pmod{3} \end{align*} and \begin{align*} 1 + \beta \equiv 0 \pmod{2} \\ 2 + 2\beta \equiv 0 \pmod{3}\\ \end{align*} And using Chinese Remainder Theorem to solve this, we get $\alpha = 4$ and $\beta = 5$, and plugging that back in yields $n = 2^6 \cdot 5^6$.

Thus, our answer is $\lfloor{ \log_{10}(2^6 \cdot 5^6) + 1 \rfloor} \Rightarrow \boxed {E}$.

~ $\color{magenta} zoomanTV$