1972 IMO Problems/Problem 1

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. $\square$

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

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