2002 AMC 10P Problems/Problem 15: Difference between revisions
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== Solution == | == Solution == | ||
There are twelve pairs <math>\{ 1, 9 \}</math>, <math>\{ 2, 10 \}</math>, <math>\{ 3, 11 \}</math>, <math>\{ 4, | There are twelve pairs <math>\{ 1, 9 \}</math>, <math>\{ 2, 10 \}</math>, <math>\{ 3, 11 \}</math>, <math>\{ 4, 12 \}</math>, <math>\; \dots \; \{ 12, 20 \}</math> in <math>\{ 1, 2, 3, \; \dots \; , 20 \}</math> that differ by 8. If we take <math>n = 12</math>, 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 <math>\boxed{\textbf{(D) } 13}</math> elements. | ||
~green_lotus | ~green_lotus | ||
Revision as of 18:46, 16 September 2024
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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