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

(solution by 4everwise/kalva, could use some further explanation of inspiration?)
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== Problem ==
 
== Problem ==
Let <math>S^{}_{}</math> be a [[subset]] of <math>\{1,2,3^{}_{},\ldots,1989\}</math> such that no two members of <math>S^{}_{}</math> differ by <math>4^{}_{}</math> or <math>7^{}_{}</math>. What is the largest number of [[element]]s <math>S^{}_{}</math> can have?
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Let <math>S</math> be a [[subset]] of <math>\{1,2,3,\ldots,1989\}</math> such that no two members of <math>S</math> differ by <math>4</math> or <math>7</math>. What is the largest number of [[element]]s <math>S</math> can have?
  
 
== Solution ==
 
== Solution ==

Revision as of 22:12, 17 July 2008

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

We first show that we can choose at most 5 numbers from $\{1, 2, \ldots , 11\}$ such that no two numbers have a difference of $4$ or $7$. We take the smallest number to be $1$, which rules out $5,8$. Now we can take at most one from each of the pairs: $[2,9]$, $[3,7]$, $[4,11]$, $[6,10]$. Now, $1989 = 180\cdot 11 + 9$, but because this isn't an exact multiple of $5$, we need to consider the last $9$ numbers.

Now let's examine $\{1, 2, \ldots , 20\}$. If we pick $1, 3, 4, 6, 9$ from the first $11$ numbers, then we're allowed to pick $11 + 1$, $11 + 3$, $11 + 4$, $11 + 6$, $11 + 9$. This means we get 10 members from the 20 numbers. Our answer is thus $179\cdot 5 + 10 = \boxed{905}$.

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