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== | ||
| − | The fact that <math>x \equiv 0 | + | The fact that <math>x \equiv 0 \mod 7 \Rightarrow 7 \mid x</math> 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. | + | 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>. |
| − | Second, note that there can be no pairs of numbers a | + | 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. |
| − | 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. | + | 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 
be a subset of 
such that no pair of distinct elements in has a sum divisible by 
. What is the maximum number of elements in ?
Solution
The fact that 
is assumed as common knowledge in this answer.
First, note that there are 
possible numbers that are equivalent to 
, and there are possible numbers equivalent to each of 
-
.
Second, note that there can be no pairs of numbers 
and 
such that 
mod , because then 
. These pairs are 
, 
, 
, and 
. Because is a pair, there can always be 
number equivalent to 
, and no more.
To maximize the amount of numbers in S, we will use number equivalent to , numbers equivalent to , and 
numbers equivalent to -
. This is obvious if you think for a moment. Therefore the answer is 
numbers.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
