Difference between revisions of "2020 AMC 10B Problems/Problem 7"
(Blanked the page) (Tag: Blanking) |
|||
| Line 1: | Line 1: | ||
| + | ==Problem== | ||
| + | How many positive even multiples of <math>3</math> less than <math>2020</math> are perfect squares? | ||
| + | |||
| + | <math>\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 12</math> | ||
| + | |||
| + | ==Solution== | ||
| + | |||
| + | 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 | ||
Revision as of 18:43, 7 February 2020
Problem
How many positive even multiples of 
less than 
are perfect squares?
Solution
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
