Difference between revisions of "2020 AMC 10B Problems/Problem 7"
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− | 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> | + | 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</math>=<math>2196 > 2020</math>, there are <math>\boxed{\textbf{(A) }7}</math> valid answers. |
~airbus-a321, November 2023 | ~airbus-a321, November 2023 | ||
Revision as of 21:59, 6 February 2024
Contents
Problem
How many positive even multiples of less than are perfect squares?
Solution 1
Any even multiple of is a multiple of , so we need to find multiples of that are perfect squares and less than . Any solution that we want will be in the form , where is a positive integer. The smallest possible value is at , and the largest is at (where the expression equals ). Therefore, there are a total of possible numbers.-PCChess
Solution 2
A even multiple square of can be represented by , where is the multiple or and makes it even. Simplifying we have . We can divide by (floor) and get see the result. We can then see that there are different values for . It can't be larger or else . And thus
~ Wiselion
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: , , , , , , and . Since =, there are valid answers. ~airbus-a321, November 2023
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
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.