2023 AMC 12A Problems/Problem 3
- The following problem is from both the 2023 AMC 10A #3 and 2023 AMC 12A #3, so both problems redirect to this page.
Contents
Problem
How many positive perfect squares less than are divisible by
?
Solution 1
Note that but
(which is over our limit of
). Therefore, the list is
. There are
elements, so the answer is
.
~zhenghua ~walmartbrian (Minor edits for clarity by Technodoggo)
Solution 2 (slightly refined)
Since , there are
perfect squares less than 2023.
~not_slay
Solution 3 (even better)
Since is prime, each solution must be divisible by
. We take
and see that there are
positive perfect squares no greater than
.
~jwseph
Solution 4
We know the highest value would be at least but less than
so we check
, prime factorizing 45. We get
. We square this and get
. We know that
, then we add 25 and get
, which does not satisfy our requirement of having the square less than
. The largest multiple of
that satisfies this is
and the smallest multiple of
that works is
so all multiples of
from
to
satisfy the requirements. Now we divide each element of the set by
and get
so there are
solutions.
~kyogrexu (minor edits by vadava_lx)
Solution 5
If we want to have the square be divisible by we must have it such that it is at least divisible by
, since every prime in it's prime factorization must have an even power.
So, we must have , and we see the range of x is $1\seq x^2\leq 8$ (Error compiling LaTeX. Unknown error_msg).
Therefore, there are solutions.
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/cMgngeSmFCY?si=E0a8wvcNRoeg2A3X&t=422
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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.