Browse · MathNet
PrintAustrian Mathematical Olympiad
Austria number theory
Problem
Anna, Berta and Clara write the square numbers on a blackboard, compute their sum and observe that it is divisible by . Then, they agree to the following game: In each round, Anna will cross out one number, then Berta will do the same, and then Clara will do the same. This continues until all numbers are crossed out. Clara has the goal that the sum of the remaining numbers after each round is divisible by .
a) Prove that Anna cannot stop Clara from reaching her goal if Clara has Berta's help.
b) Prove that Berta can stop Clara from reaching her goal even if Clara has Anna's help.
(Richard Henner)
a) Prove that Anna cannot stop Clara from reaching her goal if Clara has Berta's help.
b) Prove that Berta can stop Clara from reaching her goal even if Clara has Anna's help.
(Richard Henner)
Solution
On the blackboard, we have integers with residue modulo and integers with residue modulo . If, in a certain round, Berta and Clara cross out numbers that have the same residue modulo as the number crossed out by Anna, then they have removed either or modulo .
In both cases, the sum does not change modulo and therefore remains divisible by . Since and are multiples of , it is possible for Berta and Clara to always choose the same residue as Anna. Therefore, Anna cannot stop Clara from reaching her goal if Clara has Berta's help.
However, if Berta chooses in the first round a residue modulo that is different from the one chosen by Anna, they have crossed out or modulo . Therefore, Clara's only choices for the sum of the remaining numbers after the first round are or modulo . In both cases, the sum is not divisible by . Therefore, Clara has failed in her goal already in the first round if Berta plays uncooperatively.
(Richard Henner) ☐
In both cases, the sum does not change modulo and therefore remains divisible by . Since and are multiples of , it is possible for Berta and Clara to always choose the same residue as Anna. Therefore, Anna cannot stop Clara from reaching her goal if Clara has Berta's help.
However, if Berta chooses in the first round a residue modulo that is different from the one chosen by Anna, they have crossed out or modulo . Therefore, Clara's only choices for the sum of the remaining numbers after the first round are or modulo . In both cases, the sum is not divisible by . Therefore, Clara has failed in her goal already in the first round if Berta plays uncooperatively.
(Richard Henner) ☐
Techniques
Modular ArithmeticInvariants / monovariantsGames / greedy algorithms