2010 AMC 12B Problems/Problem 9

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

A number whose square root is a perfect cube and whose cube root is a perfect square will be in the form $a^6$ where a is an integer, because $6$ is the $LCM$ of $2$ and $3$ . A number that is divisible by $20$ obviously ends in a $0$ . The only way $a^6$ can end in a zero is if $a$ ends in a zero. The smallest number that ends in a $0$ is $10$ , so our number $a^6=10^6=1000000$ , with $\boxed {7}$ digits.

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2010 AMC 12B Problems/Problem 9

Problem 9

Solution

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