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Difference between revisions of "1992 AHSME Problems/Problem 23"

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Let <math>S</math> be a subset of <math>\{1,2,3,...,50\}</math> such that no pair of distinct elements in <math>S</math> has a sum divisible by <math>7</math>. What is the maximum number of elements in <math>S</math>?
 
Let <math>S</math> be a subset of <math>\{1,2,3,...,50\}</math> such that no pair of distinct elements in <math>S</math> has a sum divisible by <math>7</math>. What is the maximum number of elements in <math>S</math>?
 +
 +
<math>\text{(A) } 6\quad
 +
\text{(B) } 7\quad
 +
\text{(C) } 14\quad
 +
\text{(D) } 22\quad
 +
\text{(E) } 23</math>
  
 
==Solution==
 
==Solution==

Revision as of 22:09, 27 September 2014

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. $\fbox{E}$

See also

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