2020 AMC 10B Problems/Problem 7

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Problem

How many positive even multiples of $3$ less than $2020$ are perfect squares?

$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\  9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 12$

Solution

Any even multiple of $3$ is a multiple of $6$, so we need to find multiples of $6$ that are perfect squares and less than $2020$. Any solution that we want will be in the form $(6n)^2$, where $n$ is a positive integer. The smallest possible value is at $n=1$, and the largest is at $n=7$ (where the expression equals $1764$). Therefore, there are a total of $\boxed{\textbf{(A)}\ 7}$ possible numbers.-PCChess