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Difference between revisions of "2020 AMC 10B Problems/Problem 7"
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Any even multiple of <math>3</math> is a multiple of <math>6</math>, so we need to find multiples of <math>6</math> that are perfect squares and less than <math>2020</math>. Any solution that we want will be in the form <math>(6n)^2</math>, where <math>n</math> is a positive integer. The smallest possible value is at <math>n=1</math>, and the largest is at <math>n=7</math> (where the expression equals <math>1764</math>). Therefore, there are a total of <math>\boxed{\textbf{(A)}\ 7}</math> possible numbers.-PCChess
 
Any even multiple of <math>3</math> is a multiple of <math>6</math>, so we need to find multiples of <math>6</math> that are perfect squares and less than <math>2020</math>. Any solution that we want will be in the form <math>(6n)^2</math>, where <math>n</math> is a positive integer. The smallest possible value is at <math>n=1</math>, and the largest is at <math>n=7</math> (where the expression equals <math>1764</math>). Therefore, there are a total of <math>\boxed{\textbf{(A)}\ 7}</math> possible numbers.-PCChess
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==See Also==
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{{AMC10 box|year=2020|ab=A|num-b=6|num-a=8}}
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{{MAA Notice}}

Revision as of 17:44, 7 February 2020

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

See Also

2020 AMC 10A (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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