1989 AIME Problems/Problem 13

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, $1989 = 180\cdot 11 + 9$ . Because this isn't an exact multiple of $11$ , we need to consider some numbers separately.

Notice that $1969 = 180\cdot11 - 11 = 179\cdot11$ (*). Therefore we can put the last $1969$ numbers into groups of 11. 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 $179\cdot 5 + 10 = \boxed{905}$ .

Remarks (*) Suppose that you choose the values 1, 2, 4, 7, and 10. Because $1989 = 180 \times 11 + 9$ , this is not maximized. It is only maximized if we include the last element of the final set of 11, which is 10 (this is $\text{mod}(11)$ btw). To include the final element, we have to rewrite 1989 as $179 \times 11 + 20$ , which includes the final element and increases our set by 1 element.

~Brackie 1331

1989 AIME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1989 AIME Problems/Problem 13

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