2024 AMC 10A Problems/Problem 20
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[hide]Problem
Let be a subset of such that the following two conditions hold: - If and are distinct elements of , then - If and are distinct odd elements of , then . What is the maximum possible number of elements in ?
Video Solution by Scholars Foundation
https://www.youtube.com/watch?v=FKOqZau--5w&t=1s
Solution 1
All lists are organized in ascending order:
By listing out the smallest possible elements of subset we can find that subset starts with It is easily noticed that the elements of the subset "loop around" every 3 elements, specifically adding 10 each time. This means that there will be or whole loops in the subset implying that there will be elements in S. However, we have undercounted, as we did not count the remainder that resulted from With a remainder of we can fit more elements into the subset namely and resulting in a total of or elements in subset .
NOTE:
To prove that this is the best we can do, consider adding each element one by one, for the first element, say n. If n is greater than 2, we can choose n - 2 which is always better. Therefore, n = 1 or n = 2.
If n = 2 was optimal, then choose it, then the set of usable numbers in becomes 5 through 2024. We can transform the usable set of to where contains the numbers 1 through 2020. Because we assumed n = 2 was optimal, we can choose n = 2 for the set too. Because every element in is 4 below the elements of , choosing 2 in would mean choosing 6 in set . By induction we see that our list would be {2,6,10,14,18,.....2022} which only gives 506 elements which is sub-optimal. Therefore, we can conclude that n = 1 is optimal, and we proceed as the solution above.
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Video Solution by Pi Academy
https://youtu.be/fW7OGWee31c?si=oq7toGPh2QaksLHE
See also
2024 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |
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