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Difference between revisions of "1989 AIME Problems/Problem 13"

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(solution by 4everwise/kalva, could use some further explanation of inspiration?)
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== Solution ==
 
== Solution ==
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<math>S</math> can have the numbers <math>1</math> through <math>4</math>, but it can't have numbers <math>5</math> through <math>11</math>, because no two numbers can have a difference of <math>4</math> or <math>7</math>. So, <math>12</math> through <math>15</math> work, but <math>16</math> through <math>22</math> don't work, and so on. Now notice that this list contains only numbers <math>1</math> through <math>4 \mod{11}</math>. <math>1989</math> is <math>9 \mod{11}</math>, so <math>1984</math> is <math>4 \mod{11}</math>. We now have the [[sequence]]
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We first show that we can choose at most 5 numbers from <math>\{1, 2, \ldots , 11\}</math> such that no two numbers have a difference of <math>4</math> or <math>7</math>. We take the smallest number to be <math>1</math>, which rules out <math>5,8</math>. Now we can take at most one from each of the pairs: <math>[2,9]</math>, <math>[3,7]</math>, <math>[4,11]</math>, <math>[6,10]</math>. Now, <math>1989 = 180\cdot 11 + 9</math>, but because this isn't an exact multiple of <math>5</math>, we need to consider the last <math>9</math> numbers.  
  
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<cmath>\{4,15,26,...,1984\}</cmath>
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Now let's examine <math>\{1, 2, \ldots , 20\}</math>. If we pick <math>1, 3, 4, 6, 9</math> from the first <math>11</math> numbers, then we're allowed to pick <math>11 + 1</math>, <math>11 + 3</math>, <math>11 + 4</math>, <math>11 + 6</math>, <math>11 + 9</math>. This means we get 10 members from the 20 numbers. Our answer is thus <math>179\cdot 5 + 10 = \boxed{905}</math>.
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We add 7 to each term to get
 
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<cmath>\{11,22,33,...,1991\}</cmath>
 
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We divide by 11 to get
 
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<cmath>\{1,2,3,...,181\}</cmath>
 
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So there are 181 numbers <math>4 \pmod{11}</math> in S. We multiply by 4 to account for <math>1, 2</math>, and <math>3</math> <math>\mod{11}</math>: <math>181*4=\boxed{724}</math>.
 
  
 
== See also ==
 
== See also ==

Revision as of 16:52, 11 April 2008

Problem

Let $S^{}_{}$ be a subset of $\{1,2,3^{}_{},\ldots,1989\}$ such that no two members of $S^{}_{}$ differ by $4^{}_{}$ or $7^{}_{}$. What is the largest number of elements $S^{}_{}$ can have?

Solution

We first show that we can choose at most 5 numbers from $\{1, 2, \ldots , 11\}$ such that no two numbers have a difference of $4$ or $7$. We take the smallest number to be $1$, which rules out $5,8$. Now we can take at most one from each of the pairs: $[2,9]$, $[3,7]$, $[4,11]$, $[6,10]$. Now, $1989 = 180\cdot 11 + 9$, but because this isn't an exact multiple of $5$, we need to consider the last $9$ numbers.

Now let's examine $\{1, 2, \ldots , 20\}$. If we pick $1, 3, 4, 6, 9$ from the first $11$ numbers, then we're allowed to pick $11 + 1$, $11 + 3$, $11 + 4$, $11 + 6$, $11 + 9$. This means we get 10 members from the 20 numbers. Our answer is thus $179\cdot 5 + 10 = \boxed{905}$.

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
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