1992 AHSME Problems/Problem 23

==Problem== jj

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. $\fbox{E}$

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

Solution

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