Difference between revisions of "2023 AMC 12A Problems/Problem 3"
m (→Solution 6 (DO NOT DO ON AN ACTUAL TEST)) |
(→Solution 1) |
||
Line 7: | Line 7: | ||
==Solution 1== | ==Solution 1== | ||
− | Note that <math>40^2=1600</math> but <math>45^{2}=2025</math> (which is over our limit of <math>2023</math>). Therefore, the list is | + | Note that <math>40^2=1600</math> but <math>45^{2}=2025</math> (which is over our limit of <math>2023</math>). Therefore, the list is $5^2,10^2,15^ |
~zhenghua | ~zhenghua |
Revision as of 11:16, 14 April 2024
- The following problem is from both the 2023 AMC 10A #3 and 2023 AMC 12A #3, so both problems redirect to this page.
Contents
[hide]- 1 Problem
- 2 Solution 1
- 3 Solution 2 (slightly refined)
- 4 Solution 3 (the best)
- 5 Solution 4
- 6 Solution 5 (DO NOT DO ON AN ACTUAL TEST)
- 7 Video Solution by Math-X (First understand the problem!!!)
- 8 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 9 Video Solution
- 10 Video Solution (🚀 Just 2 min 🚀)
- 11 Video Solution (easy to digest) by Power Solve
- 12 See Also
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 $5^2,10^2,15^
~zhenghua ~walmartbrian (Minor edits for clarity by Technodoggo)
Solution 2 (slightly refined)
Since , there are perfect squares less than 2023.
~not_slay
Solution 3 (the best)
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 (DO NOT DO ON AN ACTUAL TEST)
Since the perfect squares have to be divisible by 5, then we know it has to be 5 times some number squared (5*x)^2. With this information, you can fique out every single product of 5 and another number squared to count how many perfect squares are divisible by 5 that are less than 2023. (EX: 5^2 = 25, 10^2 = 100, ... 40^2 = 1600) With this you get a max of 40^2, or 8 solutions. PLEASE DO NOT do this problem this way, it takes way too much time.
~BlueShardow
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/cMgngeSmFCY?si=E0a8wvcNRoeg2A3X&t=422
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=wNH6O8D-7dY
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution (🚀 Just 2 min 🚀)
~Education, the Study of Everything
Video Solution (easy to digest) by Power Solve
https://www.youtube.com/watch?v=8huvzWTtgaU
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.