Difference between revisions of "2023 AMC 10A Problems/Problem 3"
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Since <math>\left \lfloor{\sqrt{2023}}\right \rfloor = 44</math>, there are <math>\left \lfloor{\frac{44}{5}}\right \rfloor = \boxed{\textbf{(A) 8}}</math> positive integers. | Since <math>\left \lfloor{\sqrt{2023}}\right \rfloor = 44</math>, there are <math>\left \lfloor{\frac{44}{5}}\right \rfloor = \boxed{\textbf{(A) 8}}</math> positive integers. | ||
| + | |||
| + | ~not_slay | ||
==See Also== | ==See Also== | ||
Revision as of 20:50, 9 November 2023
Problem
How many positive perfect squares less than 
are divisible by 
?
![\[\textbf{(A) } 8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) } 12\]](http://latex.artofproblemsolving.com/9/6/1/961f2de34a3d109d83bde67278c21feff976ac98.png)
Solution 1
Note that 
so the list is 
there are 
elements so the answer is 
.
~zhenghua
==Solution 2 (slightly refined)
Since 
, there are 
positive integers.
~not_slay
See Also
| 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 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
