Difference between revisions of "2002 AMC 10P Problems/Problem 15"
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== Solution 2 == | == Solution 2 == | ||
+ | Use casework to find that the answer is <math>\boxed{\textbf{(D) } 13}</math>. | ||
== 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}} |
Latest revision as of 17:03, 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
Use casework to find that the answer is .
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 |
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