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

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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 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>
 +
 
~ Wiselion
 
~ Wiselion
  

Revision as of 12:20, 15 October 2023

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

~ Wiselion

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

https://www.youtube.com/watch?v=igjvQv-TCGE

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

Video Solutions

https://youtu.be/OHR_6U686Qg


https://youtu.be/5cDMRWNrH-U

~savannahsolver

Video Solution by OmegaLearn

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

~ pi_is_3.14

See Also

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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