Difference between revisions of "2020 AMC 10B Problems/Problem 7"
(→Video Solution (HOW TO CREATIVELY PROBLEM SOLVE!!!)) |
m (→Solution 2) |
||
(8 intermediate revisions by 5 users not shown) | |||
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 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>. Since <math>48^2=2196 > 2020</math>, there are <math>\boxed{\textbf{(A) }7}</math> valid answers. | ||
+ | ~airbus-a321, November 2023 | ||
==Video Solution (HOW TO CREATIVELY PROBLEM SOLVE!!!)== | ==Video Solution (HOW TO CREATIVELY PROBLEM SOLVE!!!)== |
Latest revision as of 13:18, 31 August 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 of
and
makes it even. Simplifying we have
. We can divide
by
(floor) and get
as the result. We can then see that there are
different values for
. It cannot be larger than
or else
. And thus
~ 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: ,
,
,
,
,
, 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.