1992 AHSME Problems/Problem 23

Problem

Let $S$ be a subset of $\{1,2,3,...,50\}$ such that no pair of distinct elements in $S$ has a sum divisible by $7$ . What is the maximum number of elements in $S$ ?

$\text{(A) } 6\quad \text{(B) } 7\quad \text{(C) } 14\quad \text{(D) } 22\quad \text{(E) } 23$

Solution

The fact that $x \equiv 0$ mod $7 \Rightarrow 7 \mid x$ is assumed as common knowledge in this answer.

First, note that there are 8 possible numbers that are equivalent to 1 mod 7, and there are 7 possible numbers equivalent to each of 2-6 mod 7.

Second, note that there can be no pairs of numbers a & b such that $a \equiv -b$ mod $7$ , because then a+b mod 7 = 0. These pairs are (0,0), (1,6), (2,5), and (3,4) mod 7. Because (0,0) is a pair, there can always be 1 number equivalent to 0 mod 7, and no more.

To maximize the amount of numbers in S, we will use 1 number equivalent to 0 mod 7, 8 numbers equivalent to 1, and 14 numbers equivalent to 2-5. This is obvious if you think for a moment. Therefore the answer is 1+8+14=23 numbers. $\fbox{E}$

1992 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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1992 AHSME Problems/Problem 23

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