Difference between revisions of "2023 AMC 12A Problems/Problem 3"
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==Solution 1== | ==Solution 1== | ||
| − | Note that <math>45^{2}=2025</math> so the list is <math>5,10,15,20,25,30,35,40</math> | + | Note that <math>40^2=1600</math> but <math>45^{2}=2025</math> (which is over our limit of <math>2023</math>) so the list is <math>5^2,10^2,15^2,20^2,25^2,30^2,35^2,40^2</math>. There are <math>8</math> elements, so the answer is <math>\boxed{\textbf{(A) 8}}</math>. |
~zhenghua | ~zhenghua | ||
| + | |||
| + | (Minor edits for clarity by Technodoggo) | ||
==Solution 2 (slightly refined)== | ==Solution 2 (slightly refined)== | ||
Revision as of 23:50, 9 November 2023
Problem
How many positive perfect squares less than 
are divisible by 
?
Solution 1
Note that 
but 
(which is over our limit of ) so the list is 
. There are 
elements, so the answer is 
.
~zhenghua
(Minor edits for clarity by Technodoggo)
Solution 2 (slightly refined)
Since 
, there are 
perfect squares less than 2023.
~not_slay
See Also
| 2023 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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 | ||
| 2023 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 2 |
Followed by Problem 4 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
