1972 IMO Problems/Problem 1

Revision as of 15:25, 10 November 2023 by Ehuang0531 (talk | contribs) (technicality - assuming A and B must be nonempty)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)


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.


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