1989 AIME Problems/Problem 13
Problem
Let be a subset of such that no two members of differ by or . What is the largest number of elements can have?
Solution
S can have the numbers 1 through 4, but it can't have numbers 5 through 11, because no two numbers can have a difference of 4 or 7. So, 12 through 15 work, but 16 through 22 don't work. So on. Now notice that this list contains only numbers 1 through 4 mod 11. 1989 is 9 mod 11, so 1984 is 4 mod 11. We now have the sequence
{4,15,26,...,1984}
We add 7 to each term to get
{11,22,33,...,1991}
We divide by 11 to get
{1,2,3,...,181}
So there are 181 numbers 4 mod 11 in S. We multiply by 4 to account for 1, 2, and 3 mod 11:
See also
1989 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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