Difference between revisions of "2023 AMC 12A Problems/Problem 3"
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We know the highest value would be at least <math>40</math> but less than <math>50</math> so we check <math>45</math>, prime factorizing 45. We get <math>3^2 \cdot 5</math>. We square this and get <math>81 \cdot 25</math>. We know that <math>80 \cdot 25 = 2000</math>, then we add 25 and get <math>2025</math>, which does not satisfy our requirement of having the square less than <math>2023</math>. The largest multiple of <math>5</math> that satisfies this is <math>40</math> and the smallest multiple of <math>5</math> that works is <math>5</math> so all multiples of <math>5</math> from <math>5</math> to <math>40</math> satisfy the requirements. Now we divide each element of the set by <math>5</math> and get <math>1-8</math> so there are <math>\boxed{\textbf{(A) 8}}</math> solutions. | We know the highest value would be at least <math>40</math> but less than <math>50</math> so we check <math>45</math>, prime factorizing 45. We get <math>3^2 \cdot 5</math>. We square this and get <math>81 \cdot 25</math>. We know that <math>80 \cdot 25 = 2000</math>, then we add 25 and get <math>2025</math>, which does not satisfy our requirement of having the square less than <math>2023</math>. The largest multiple of <math>5</math> that satisfies this is <math>40</math> and the smallest multiple of <math>5</math> that works is <math>5</math> so all multiples of <math>5</math> from <math>5</math> to <math>40</math> satisfy the requirements. Now we divide each element of the set by <math>5</math> and get <math>1-8</math> so there are <math>\boxed{\textbf{(A) 8}}</math> solutions. | ||
− | ~kyogrexu | + | ~kyogrexu(minor edits by vadava_lx) |
− | |||
==Video Solution by Math-X (First understand the problem!!!)== | ==Video Solution by Math-X (First understand the problem!!!)== |
Revision as of 21:31, 10 November 2023
- 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]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)
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.