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Difference between revisions of "2010 AMC 12B Problems/Problem 9"

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== Problem 9 ==
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== Problem ==
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Let <math>n</math> be the smallest positive integer such that <math>n</math> id divisible by <math>20</math>, <math>n^2</math> is a perfect cube, and <math>n^3</math> is a perfect square. What is the number of digits of <math>n</math>?
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Let <math>n</math> be the smallest positive integer such that <math>n</math> is divisible by <math>20</math>, <math>n^2</math> is a perfect cube, and <math>n^3</math> is a perfect square. What is the number of digits of <math>n</math>?
  
 
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7</math>
 
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7</math>
  
 
== Solution ==
 
== Solution ==
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We know that <math>n^2 = k^3</math> and <math> n^3 = m^2 </math>. Cubing and squaring the equalities respectively gives <math> n^6 = k^9 = m^4 </math>. Let <math>a = n^6</math>. Now we know <math>a</math> must be a perfect <math>36</math>-th power because <math>lcm(9,4) = 36</math>, which means that <math>n</math> must be a perfect <math>6</math>-th power. The smallest number whose sixth power is a multiple of <math>20</math> is <math>10</math>, because the only prime factors of <math>20</math> are <math>2</math> and <math>5</math>, and <math>10 = 2 \times 5</math>. Therefore our is equal to number <math>10^6 = 1000000</math>, with <math>7</math> digits <math>\Rightarrow \boxed {E}</math>.
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 +
== Solution 2 (Chinese Remainder Theorem) ==
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Let <math>n = 2^2 \cdot 5 \cdot x</math> for some integer <math>x</math>, then we know that
 +
\begin{align*}
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n^2 = 2^4 \cdot 5^2 \cdot x^2 = m_0^3\\
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n^3 = 2^6 \cdot 5^3 \cdot x^3 = m_1^2
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\end{align*}
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Since we have <math>2^4 \cdot 5^2 \cdot x^2</math> 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 <math>p \mid x</math> if <math>p \not \in \{2,3\}</math>, since that would just serve to increase <math>n</math> and we want <math>n</math> to be minimal.
 +
 +
This means we can write <math>x = 2^{\alpha} \cdot 5^{\beta}</math> and thus we have
 +
<math>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</math>, therefore we know
 +
\begin{align*}
 +
4 + 2\alpha \equiv 0 \pmod{3}\\
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2 + 2\beta \equiv 0 \pmod{3}
 +
\end{align*}
 +
and doing the same with the third equation being a square, we have
 +
\begin{align*}
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6 + 3\alpha \equiv 0 \pmod{2}\\
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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*}
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And using [[Chinese Remainder Theorem]] to solve this, we get <math>\alpha = 4</math> and <math>\beta = 5</math>, and plugging that back in yields <math>n = 2^6 \cdot 5^6</math>.
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 +
Thus, our answer is <math>\lfloor{ \log_{10}(2^6 \cdot 5^6) + 1 \rfloor} \Rightarrow \boxed {E}</math>.
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 +
~ <math>\color{magenta} zoomanTV</math>
  
 
== See also ==
 
== See also ==
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{{AMC12 box|year=2010|num-b=12|num-a=14|ab=B}}
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{{AMC12 box|year=2010|num-b=8|num-a=10|ab=B}}
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{{MAA Notice}}

Latest revision as of 14:52, 6 May 2024

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

~ $\color{magenta} zoomanTV$

See also

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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All AMC 12 Problems and Solutions

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