Difference between revisions of "1989 AIME Problems/Problem 13"

m (Solution)
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== Solution ==
 
== 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
+
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}
 
{4,15,26,...,1984}
  
we add 7 to each term to get
+
We add 7 to each term to get
  
 
{11,22,33,...,1991}
 
{11,22,33,...,1991}

Revision as of 13:29, 15 October 2007

Problem

Let $S^{}_{}$ be a subset of $\{1,2,3^{}_{},\ldots,1989\}$ such that no two members of $S^{}_{}$ differ by $4^{}_{}$ or $7^{}_{}$. What is the largest number of elements $S^{}_{}$ 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:

$181*4=\boxed{724}$

See also

1989 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions