Difference between revisions of "1989 AIME Problems/Problem 13"
m |
|||
| Line 12: | Line 12: | ||
[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] | ||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 18:17, 4 July 2013
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
We first show that we can choose at most 5 numbers from 
such that no two numbers have a difference of or . We take the smallest number to be 
, which rules out 
. Now we can take at most one from each of the pairs: ![$[2,9]$](http://latex.artofproblemsolving.com/7/2/c/72c5cb8ada1854b975804507b0666d15f38b8f84.png)
, ![$[3,7]$](http://latex.artofproblemsolving.com/2/4/1/241f0fcc823cd1de486bae2259885a82409237c5.png)
, ![$[4,11]$](http://latex.artofproblemsolving.com/9/3/e/93ef02ac376b1998d0cee938f2d41b6cc1d49c50.png)
, ![$[6,10]$](http://latex.artofproblemsolving.com/1/e/0/1e0d5b8f6601c37464a770768f132bf1162ecc38.png)
. Now, 
, but because this isn't an exact multiple of 
, we need to consider the last 
numbers.
Now let's examine 
. If we pick 
from the first 
numbers, then we're allowed to pick 
, 
, 
, 
, 
. This means we get 10 members from the 20 numbers. Our answer is thus 
.
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. 
