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

(Solution (Cheap Way))
 
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==Solution==
 
==Solution==
  
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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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>91+92+93+94+95+96+97+98+99</math>, which is less than <math>100+100+100+100+100+100+100+100+100=1000 < 1022</math>. There are even less 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>\square</math>
 
 
==Solution (Cheap Way)==
 
 
 
By definition an empty set is disjoint to any other set. Therefore, our subsets will be empty set and empty set.
 
 
 
{{alternate solutions}}
 
  
 
==See Also==
 
==See Also==

Latest revision as of 20:31, 12 March 2018

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

Note that there are $2^{10}-2=1022$ 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 $91+92+93+94+95+96+97+98+99$, which is less than $100+100+100+100+100+100+100+100+100=1000 < 1022$. There are even less attainable sums. The Pigeonhole Principle then implies that there are two distinct subsets whose members have the same sum. Let these sets be $A$ and $B$. Note that $A- (A\cap B)$ and $B- (A\cap B)$ 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. $\square$

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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