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

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==Solution==
 
==Solution==
  
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The fact that <math>x \equiv 0</math> mod <math>7 \Rightarrow 7 \mid x</math> is assumed as common knowledge in this answer.
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The fact that <math>x \equiv 0 \mod 7 \Rightarrow 7 \mid x</math> is assumed as common knowledge in this answer.
  
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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.
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First, note that there are <math>8</math> possible numbers that are equivalent to <math>1 \mod 7</math>, and there are <math>7</math> possible numbers equivalent to each of <math>2</math>-<math>6 \mod 7</math>.
  
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Second, note that there can be no pairs of numbers a & b such that <math>a \equiv -b</math> mod <math>7</math>, 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.
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Second, note that there can be no pairs of numbers <math>a</math> and <math>b</math> such that <math>a \equiv -b</math> mod <math>7</math>, because then <math>a+b | 7</math>. These pairs are <math>(0,0)</math>, <math>(1,6)</math>, <math>(2,5)</math>, and <math>(3,4)</math>. Because <math>(0,0)</math> is a pair, there can always be <math>1</math> number equivalent to <math>0 \mod 7</math>, and no more.
  
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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.
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To maximize the amount of numbers in S, we will use <math>1</math> number equivalent to <math>0 \mod 7</math>, <math>8</math> numbers equivalent to <math>1</math>, and <math>14</math> numbers equivalent to <math>2</math>-<math>5</math>. This is obvious if you think for a moment. Therefore the answer is <math>1+8+14=23</math> numbers.
 
<math>\fbox{E}</math>
 
<math>\fbox{E}</math>
  

Latest revision as of 12:25, 16 July 2024

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 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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