Difference between revisions of "1989 AIME Problems/Problem 13"
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== Solution == | == Solution == | ||
| − | {{ | + | S can have the numbers 1 through 4, but it can't have numbers 5 through 11, because no two numbers can have a difference of 4 or 7. so 12 through 15 work, but 16 through 22 don't work. So on. Now notice that this list contains only numbers 1 through 4 mod 11. 1989 is 9 mod 11, so 1984 is 4 mod 11. We now have the sequence |
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| + | {4,15,26,...,1984} | ||
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| + | we add 7 to each term to get | ||
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| + | {11,22,33,...,1991} | ||
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| + | We divide by 11 to get | ||
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| + | {1,2,3,...,181} | ||
| + | |||
| + | So there are 181 numbers 4 mod 11 in S. We multiply by 4 to account for 1, 2, and 3 mod 11: | ||
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| + | <math>181*4=\boxed{724}</math> | ||
== See also == | == See also == | ||
| − | + | {{AIME box|year=1989|num-b=12|num-a=14}} | |
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Revision as of 07:48, 15 October 2007
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
S can have the numbers 1 through 4, but it can't have numbers 5 through 11, because no two numbers can have a difference of 4 or 7. so 12 through 15 work, but 16 through 22 don't work. So on. Now notice that this list contains only numbers 1 through 4 mod 11. 1989 is 9 mod 11, so 1984 is 4 mod 11. We now have the sequence
{4,15,26,...,1984}
we add 7 to each term to get
{11,22,33,...,1991}
We divide by 11 to get
{1,2,3,...,181}
So there are 181 numbers 4 mod 11 in S. We multiply by 4 to account for 1, 2, and 3 mod 11:
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 | ||
