# 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 $a$ and $b$ such that $a \equiv -b$ mod $7$, because then $a+b | 7$. These pairs are $(0,0)$, $(1,6)$, $(2,5)$, and $(3,4)$. 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. $\fbox{E}$

## See also

 1992 AHSME (Problems • Answer Key • Resources) Preceded byProblem 22 Followed byProblem 24 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 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

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