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 <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 <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 <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> | ||
− | + | == See also == | |
+ | {{AHSME box|year=1992|num-b=22|num-a=24}} | ||
− | + | [[Category: Intermediate Number Theory Problems]] | |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 13:25, 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 |
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