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Difference between revisions of "1972 IMO Problems/Problem 1"

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Prove that from a set of ten distinct two-digit numbers (in the decimal system), it is possible to select two disjoint subsets whose members have the same sum.  
 
Prove that from a set of ten distinct two-digit numbers (in the decimal system), it is possible to select two disjoint subsets whose members have the same sum.  
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==Solution==
 
==Solution==
  
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Note that there are <math>2^{10}-2=1022</math> distinct subsets of our set of 10 two-digit numbers. Also note that the sum of the elements of any subset of our set of 10 two-digit numbers must be between 10 and <math>90+91+92+93+94+95+96+97+98+99=945</math>. This shows that there are 846 attainable sums. The [[Pigeonhole Principle]] then implies that there are two distinct subsets whose members have the same sum. Let these sets be <math>A</math> and <math>B</math>. Note that <math>A- (A\cap B)</math> and <math>B- (A\cap B)</math> are two distinct sets whose members have the same sum. These two sets are subsets of our set of 10 distinct two-digit numbers, so this proves the claim. <math>\blacksquare</math>
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There are <math>2^{10}-2=1022</math> distinct subsets of our set of 10 two-digit numbers. The sum of the elements of any subset of our set of 10 two-digit numbers must be between <math>10</math> and <math>90+91+92+93+94+95+96+97+98+99 < 10 \cdot 100 = 1000</math>. (There are fewer attainable sums.) As <math>1000 < 1022</math>, the [[Pigeonhole Principle]] implies there are two distinct subsets whose members have the same sum. Let these sets be <math>A</math> and <math>B</math>. Now, let <math>A' = A - (A \cap B)</math> and <math>B' = B - (A \cap B)</math>. Notice <math>A'</math> and <math>B'</math> are disjoint. They are also nonempty because if <math>A = A \cap B</math> or <math>B = A \cap B</math>, then one of <math>A</math> and <math>B</math> is a subset of the other, so they are either not distinct or have different sums. Therefore <math>A'</math> and <math>B'</math> are disjoint subsets our set of 10 distinct two-digit numbers, which proves the claim. <math>\square</math>
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{{alternate solutions}}
 
  
 
==See Also==
 
==See Also==

Latest revision as of 15:25, 10 November 2023

Problem

Prove that from a set of ten distinct two-digit numbers (in the decimal system), it is possible to select two disjoint subsets whose members have the same sum.

Solution

There are $2^{10}-2=1022$ distinct subsets of our set of 10 two-digit numbers. The sum of the elements of any subset of our set of 10 two-digit numbers must be between $10$ and $90+91+92+93+94+95+96+97+98+99 < 10 \cdot 100 = 1000$. (There are fewer attainable sums.) As $1000 < 1022$, the Pigeonhole Principle implies there are two distinct subsets whose members have the same sum. Let these sets be $A$ and $B$. Now, let $A' = A - (A \cap B)$ and $B' = B - (A \cap B)$. Notice $A'$ and $B'$ are disjoint. They are also nonempty because if $A = A \cap B$ or $B = A \cap B$, then one of $A$ and $B$ is a subset of the other, so they are either not distinct or have different sums. Therefore $A'$ and $B'$ are disjoint subsets our set of 10 distinct two-digit numbers, which proves the claim. $\square$

See Also

1972 IMO (Problems) • Resources
Preceded by
First Problem 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
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