2020 AMC 10B Problem 7

P1-10~2 minPrint

Number Properties Divisibility & Factors Bound the quantity above/below or estimate to pin it down

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

Show solution without answering

Key insight

A square that is a multiple of 6 must be the square of a multiple of 6, so list 6 squared, 12 squared, and so on below 2020.

Solution

An even multiple of $3$ is a number divisible by both $2$ and $3$ , that is, a multiple of $6$ .
Suppose a perfect square $m \times m$ is a multiple of $6$ . Then $m$ itself must be even: if $m$ were odd, then $m \times m$ would be odd. In the same way $m$ must be a multiple of $3$ , because $3$ is prime, so it can only divide $m \times m$ if it divides $m$ . So $m$ is a multiple of $6$ .
Now list the squares of multiples of $6$ and stop when they pass $2020$ :

$m$	$6$	$12$	$18$	$24$	$30$	$36$	$42$	$48$
$m^2$	$36$	$144$	$324$	$576$	$900$	$1296$	$1764$	$2304$

The first seven are below $2020$ . The eighth, $2304$ , is too big, and the squares only get bigger after that.

So there are $7$ such squares.

The answer is $\boxed{\textbf{(A)}\ 7}$ .

Why this works

If a prime divides a perfect square, it divides the number being squared. That turns "squares that are multiples of $6$ " into "squares of multiples of $6$ ", which are easy to list. You can also shortcut the list: the squares are $36 \times 1, 36 \times 4, 36 \times 9, \ldots$ , and $36 \times 49 = 1764 < 2020 < 2304 = 36 \times 64$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

An even multiple of $3$ is a multiple of $6$ . If a perfect square $m^2$ is divisible by $2$ and by $3$ , then $m$ itself is divisible by $2$ and $3$ , so $m = 6k$ and $m^2 = 36k^2$ .

We need $36k^2 < 2020$ , i.e. $k^2 < \tfrac{2020}{36} \approx 56.1$ . Since $7^2 = 49 < 56.1 < 64 = 8^2$ , the valid values are $k = 1, 2, \dots, 7$ .

Check the boundary: $36 \cdot 49 = 1764 < 2020$ and $36 \cdot 64 = 2304 > 2020$ .

The answer is $\boxed{\textbf{(A)}\ 7}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Algebra, Chapter 20: Special Functions (§20.3)
Introduction to Number Theory, Chapter 5: Divisor Problems (§5.2, §5.3)
Introduction to Number Theory, Chapter 13: Divisibility Rules (§13.2)