Difference between revisions of "2020 AMC 10B Problems/Problem 7"

Latest revision as of 19:44, 2 November 2024

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 1

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

Solution 2

A even multiple square of $3$ can be represented by $3^2 \cdot 2^2 \cdot x^2$ , where $3^2$ is the multiple of $3$ and $2^2$ makes it even. Simplifying we have $36 \cdot x^2$ . We can divide $2020$ by $36$ (floor) and get $56$ as the result. We can then see that there are $7$ different values for $x$ . It cannot be larger than $7$ or else $x^2 > 56$ . And thus $\boxed{\textbf{(A) }7}$

~ Wiselion

~<B+

Solution 3

It can be seen that the problem is just asking for squares that are multiples of six. Thus, all squares of multiples of six can be listed out: $6^2$ , $12^2$ , $18^2$ , $24^2$ , $30^2$ , $36^2$ , and $42^2$ . Since $48^2=2304 > 2020$ and $42^2=1764 < 2020$ , there are $\boxed{\textbf{(A) }7}$ valid answers. ~airbus-a321, November 2023

Video Solution (HOW TO CREATIVELY PROBLEM SOLVE!!!)

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Check It Out! Short & Straight-Forward Solution ~Education, The Study of Everything

Video Solutions

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Video Solution by OmegaLearn

https://youtu.be/ZhAZ1oPe5Ds?t=2241

~ pi_is_3.14

@@ Line 11: / Line 11: @@
 ==Solution 2==
-A even multiple square of <math>3</math> can be represented by <math>3^2 \cdot 2^2 \cdot x^2</math>, where <math>3^2</math> is the multiple or <math>3</math> and <math>2^2</math> makes it even. Simplifying we have <math>36^2 \cdot x^2</math>. We can divide <math>2020</math> by <math>36</math> (floor) and get <math>56</math> see the result. We can then see that there are <math>7</math> different values for <math>x</math>. It can't be larger or else <math>x^2 > 56</math>. And thus <math>\boxed{\textbf{(A) }7}</math>
+A even multiple square of <math>3</math> can be represented by <math>3^2 \cdot 2^2 \cdot x^2</math>, where <math>3^2</math> is the multiple of <math>3</math> and <math>2^2</math> makes it even. Simplifying we have <math>36 \cdot x^2</math>. We can divide <math>2020</math> by <math>36</math> (floor) and get <math>56</math> as the result. We can then see that there are <math>7</math> different values for <math>x</math>. It cannot be larger than <math>7</math> or else <math>x^2 > 56</math>. And thus <math>\boxed{\textbf{(A) }7}</math>
 ~ Wiselion
+~<B+
 ==Solution 3==
-It can be seen that the problem is just asking for squares that are multiples of six. Thus, all squares of multiples of six can be listed out: <math>6^2</math>, <math>12^2</math>, <math>18^2</math>, <math>24^2</math>, <math>30^2</math>, <math>36^2</math>, and <math>42^2</math>. <math>48^2</math>=<math>2196 > 2020</math>. There are seven valid answers.
+It can be seen that the problem is just asking for squares that are multiples of six. Thus, all squares of multiples of six can be listed out: <math>6^2</math>, <math>12^2</math>, <math>18^2</math>, <math>24^2</math>, <math>30^2</math>, <math>36^2</math>, and <math>42^2</math>. Since <math>48^2=2304 > 2020</math> and <math>42^2=1764 < 2020</math>, there are <math>\boxed{\textbf{(A) }7}</math> valid answers.
 ~airbus-a321, November 2023

2020 AMC 10B (Problems • Answer Key • Resources)
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