Difference between revisions of "2020 AMC 10B Problems/Problem 7"
(→Video Solution (HOW TO CREATIVELY PROBLEM SOLVE!!!)) |
(→Solution) |
||
| Line 5: | Line 5: | ||
<math>\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 12</math> | <math>\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 12</math> | ||
| − | ==Solution== | + | ==Solution 1== |
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 | ||
| + | |||
| + | == 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> | ||
| + | ~ Wiselion | ||
==Video Solution (HOW TO CREATIVELY PROBLEM SOLVE!!!)== | ==Video Solution (HOW TO CREATIVELY PROBLEM SOLVE!!!)== | ||
Revision as of 11:20, 15 October 2023
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
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. 
