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

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

~ $\color{magenta} zoomanTV$

2010 AMC 12B (Problems • Answer Key • Resources)
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

Page

Toolbox

Search

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

Latest revision as of 14:52, 6 May 2024

Contents

Problem

Solution

Solution 2 (Chinese Remainder Theorem)

See also

@@ Line 1: / Line 1: @@
-== Problem 9 ==
+== Problem ==
 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>?
@@ Line 5: / Line 5: @@
 == Solution ==
+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>.
+== Solution 2 (Chinese Remainder Theorem) ==
+Let <math>n = 2^2 \cdot 5 \cdot x</math> for some integer <math>x</math>, 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 <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*}
++ 2\alpha \equiv 0 \pmod{3}\\
++ 2\beta \equiv 0 \pmod{3}
+\end{align*}
+and doing the same with the third equation being a square, we have
+\begin{align*}
++ 3\alpha \equiv 0 \pmod{2}\\
++ 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}\\
++ 2\alpha \equiv 0 \pmod{3}
+\end{align*}
+and
+\begin{align*}
++ \beta \equiv 0 \pmod{2} \\
++ 2\beta \equiv 0 \pmod{3}\\
+\end{align*}
+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>.
+Thus, our answer is <math>\lfloor{ \log_{10}(2^6 \cdot 5^6) + 1 \rfloor} \Rightarrow \boxed {E}</math>.
+~ <math>\color{magenta} zoomanTV</math>
 == See also ==
 {{AMC12 box|year=2010|num-b=8|num-a=10|ab=B}}
+{{MAA Notice}}