2012 AMC 12A Problem 17

Attempt Problem 17 of the 2012 AMC 12A 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 2012 AMC 12A solutions, or check the answer key.

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17.

Let SS be a subset of {1,2,3,,30}\{1, 2, 3, \ldots, 30\} with the property that no pair of distinct elements in SS has a sum divisible by 5.5. What is the largest possible size of S?S?

1010

1313

1515

1616

1818

Answer: B
Concepts:modular arithmeticsubsetsextremal argument
Difficulty rating: 1800
Solution:

Group {1,,30}\{1, \ldots, 30\} by residue modulo 5;5; each class has 66 numbers. A sum is divisible by 55 when the residues are 0+0,0{+}0, 1+4,1{+}4, or 2+3.2{+}3.

So SS can use at most one number 0,\equiv 0, and only one of the classes {1},{4}\{1\}, \{4\} and only one of {2},{3}.\{2\}, \{3\}. That allows at most 1+6+6=131 + 6 + 6 = 13 numbers.

The set {1,2,6,7,11,12,\{1, 2, 6, 7, 11, 12, 16,17,21,22,16, 17, 21, 22, 26,27,30}26, 27, 30\} achieves 13,13, so the maximum is 13.13.

Thus, the correct answer is B.

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