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

Revision as of 09:59, 4 July 2013

Problem 9

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 $\Rightarrow \boxed {E}$ .

2010 AMC 12B (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 == See also ==
 {{AMC12 box|year=2010|num-b=8|num-a=10|ab=B}}
+{{MAA Notice}}

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

Revision as of 09:59, 4 July 2013

Problem 9

Solution

See also