Difference between revisions of "1986 AIME Problems/Problem 12"

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Let the sum of a set of numbers be the sum of its elements. Let <math>\displaystyle S</math> be a set of positive integers, none greater than 15. Suppose no two disjoint subsets of <math>\displaystyle S</math> have the same sum. What is the largest sum a set <math>\displaystyle S</math> with these properties can have?  
 
Let the sum of a set of numbers be the sum of its elements. Let <math>\displaystyle S</math> be a set of positive integers, none greater than 15. Suppose no two disjoint subsets of <math>\displaystyle S</math> have the same sum. What is the largest sum a set <math>\displaystyle S</math> with these properties can have?  
 
== Solution ==
 
== Solution ==
{{solution}}
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The maximum is <math>61</math>, attained when <math>S=\{ 15,14,13,11,8\}</math>. We must now prove that no such set has sum at least 62. Suppose such a set <math>S</math> existed. Then <math>S</math> can't have 4 or less elements, otherwise its sum would be at most <math>15+14+13+12=54</math>.
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But also, <math>S</math> can't have at least 6 elements. To see why, note that <math>2^6-1-1-6=56</math> of its subsets have at most four elements, so each of them have sum at most 54. By the Pigeonhole Principle, two of these subsets would have the same sum, contradiction.
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It follows that <math>S</math> must have exactly 5 elements. <math>S</math> contains both 15 and 14 (otherwise its sum is at most <math>10+11+12+13+15=61</math>). It follows that <math>S</math> cannot contain both <math>a</math> and <math>a-1</math> for any <math>a\leq 13</math>. So now <math>S</math> must contain 13 (otherwise its sum is at most <math>15+14+12+10+8=59</math>), and <math>S</math> cannot contain 12.
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Now the only way <math>S</math> could have sum at least <math>62=15+14+13+11+9</math> would be if <math>S=\{ 15,14,13,11,9\}</math>. But <math>15+11=13+9</math> so this set does not work, a contradiction. Therefore 61 is indeed the maximum.
 
== See also ==
 
== See also ==
 
* [[1986 AIME Problems]]
 
* [[1986 AIME Problems]]
  
 
{{AIME box|year=1986|num-b=11|num-a=13}}
 
{{AIME box|year=1986|num-b=11|num-a=13}}

Revision as of 20:18, 26 March 2007

Problem

Let the sum of a set of numbers be the sum of its elements. Let $\displaystyle S$ be a set of positive integers, none greater than 15. Suppose no two disjoint subsets of $\displaystyle S$ have the same sum. What is the largest sum a set $\displaystyle S$ with these properties can have?

Solution

The maximum is $61$, attained when $S=\{ 15,14,13,11,8\}$. We must now prove that no such set has sum at least 62. Suppose such a set $S$ existed. Then $S$ can't have 4 or less elements, otherwise its sum would be at most $15+14+13+12=54$.

But also, $S$ can't have at least 6 elements. To see why, note that $2^6-1-1-6=56$ of its subsets have at most four elements, so each of them have sum at most 54. By the Pigeonhole Principle, two of these subsets would have the same sum, contradiction.


It follows that $S$ must have exactly 5 elements. $S$ contains both 15 and 14 (otherwise its sum is at most $10+11+12+13+15=61$). It follows that $S$ cannot contain both $a$ and $a-1$ for any $a\leq 13$. So now $S$ must contain 13 (otherwise its sum is at most $15+14+12+10+8=59$), and $S$ cannot contain 12.

Now the only way $S$ could have sum at least $62=15+14+13+11+9$ would be if $S=\{ 15,14,13,11,9\}$. But $15+11=13+9$ so this set does not work, a contradiction. Therefore 61 is indeed the maximum.

See also

1986 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AIME Problems and Solutions
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