2024 AMC 10A Problem 20

Attempt Problem 20 of the 2024 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 10A solutions, or check the answer key.

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20.

Let SS be a subset of {1,2,3,,2024}\{1, 2, 3, \ldots, 2024\} such that the following two conditions hold:

• If xx and yy are distinct elements of S,S, then xy>2.|x - y| \gt 2.

• If xx and yy are distinct odd elements of S,S, then xy>6.|x - y| \gt 6.

What is the maximum possible number of elements in S?S?

436436

506506

608608

654654

675675

Answer: C
Concepts:arrangements with restrictionsextremal argumentoptimization
Difficulty rating: 2080
Solution:

The two conditions say chosen numbers are at least 33 apart, and chosen odd numbers at least 77 apart. Any four numbers in a block of 1010 would need three gaps of at least 3,3, so they would have to occupy positions r,r+3,r+6,r+9.r,r+3,r+6,r+9. Two of the odd entries would then differ by 6,6, which is forbidden. Thus each full block of 1010 contains at most 33 choices, and the last four positions contain at most 2.2. This gives the upper bound 2023+2=608.202\cdot3+2=608. It is attained by the pattern 1,4,8,11,14,18,1,4,8,11,14,18,\ldots (residues 1,4,8(mod10)1,4,8\pmod{10}), together with 2021,2024.2021,2024. Adjacent selected values differ by at least 3,3, and the selected odd values are 1010 apart. Therefore, the answer is C.

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