Difference between revisions of "1992 AHSME Problems/Problem 23"
| Line 12: | Line 12: | ||
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. | ||
| + | {{MAA Notice}} | ||
Revision as of 12:55, 5 July 2013
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.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
