Difference between revisions of "2023 AMC 10A Problems/Problem 3"
Arcticturn (talk | contribs) (→Solution 1) |
(→Solution 1) |
||
Line 9: | Line 9: | ||
~zhenghua | ~zhenghua | ||
+ | |||
+ | ==Solution 2== | ||
+ | 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. | ||
==See Also== | ==See Also== |
Revision as of 20:50, 9 November 2023
Contents
Problem
How many positive perfect squares less than are divisible by ?
Solution 1
Note that so the list is there are elements so the answer is .
~zhenghua
Solution 2
Since , there are positive integers.
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.