Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
 
Line 9: Line 9:
 
Remarks
 
Remarks
 
(*) Suppose that you choose the values 1, 2, 4, 7, and 10. Because <math>1989 = 180 \times 11 + 9</math>, 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 <math>\text{mod}(11)</math> btw). To include the final element, we have to rewrite 1989 as <math>179 \times 11 + 20</math>, which includes the final element and increases our set by 1 element.
 
(*) Suppose that you choose the values 1, 2, 4, 7, and 10. Because <math>1989 = 180 \times 11 + 9</math>, 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 <math>\text{mod}(11)</math> btw). To include the final element, we have to rewrite 1989 as <math>179 \times 11 + 20</math>, which includes the final element and increases our set by 1 element.
 +
 +
~Brackie 1331
  
 
== See also ==
 
== See also ==

Latest revision as of 03:07, 17 December 2023

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

See also

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
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_13&oldid=207752"