2010 AMC 12B Problems/Problem 9
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 \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 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 if , since that would just serve to increase and we want to be minimal.
This means we can write and thus we have , 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 and , and plugging that back in yields .
Thus, our answer is .
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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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