Difference between revisions of "2002 AMC 10P Problems/Problem 15"
m (→Solution: linking pigeonhole principle) |
(→Solution) |
||
| Line 8: | Line 8: | ||
~green_lotus | ~green_lotus | ||
| + | |||
| + | |||
| + | |||
| + | == Solution 2 == | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2002|ab=P|num-b=14|num-a=16}} | {{AMC10 box|year=2002|ab=P|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:03, 14 August 2024
Contents
Problem
What is the smallest integer 
for which any subset of 
of size must contain two numbers that differ by 8?
Solution
There are twelve pairs 
, 
, 
, 
, 
in that differ by 8. If we take 
, it could be that we selected one element from each pair for the subset: the condition may not be fulfilled. By the Pigeonhole principle, in order to select at least one pair, it is necessary to select 
elements.
~green_lotus
Solution 2
See Also
| 2002 AMC 10P (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
