1972 IMO Problems/Problem 1

Revision as of 19:31, 12 March 2018 by Tree3 (talk | contribs) (Solution (Cheap Way))
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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
Invalid username
Login to AoPS