1992 AHSME Problems/Problem 23
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. $\fbox{E}
See also
1992 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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