Difference between revisions of "1992 AHSME Problems/Problem 23"
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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. | 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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== See also == | == See also == | ||
Revision as of 16:04, 27 September 2014
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 
mod 
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 
mod , 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.
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. 
