Difference between revisions of "2010 AMC 12B Problems/Problem 9"
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− | == Problem | + | == 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>? | 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>? | ||
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== Solution == | == 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*} | ||
+ | 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 <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 == | == See also == | ||
− | {{AMC12 box|year=2010|num-b= | + | {{AMC12 box|year=2010|num-b=8|num-a=10|ab=B}} |
+ | {{MAA Notice}} |
Latest revision as of 14:52, 6 May 2024
Problem
Let be the smallest positive integer such that is divisible by , is a perfect cube, and is a perfect square. What is the number of digits of ?
Solution
We know that and . Cubing and squaring the equalities respectively gives . Let . Now we know must be a perfect -th power because , which means that must be a perfect -th power. The smallest number whose sixth power is a multiple of is , because the only prime factors of are and , and . Therefore our is equal to number , with digits .
Solution 2 (Chinese Remainder Theorem)
Let for some integer , then we know that
This means we can write and thus we have
, therefore we know
Thus, our answer is .
~
See also
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 |
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