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Problem 1548

National Olympiad, first round

Combinatorics Difficulty 6.0 Prove it Estonian Mathematical Olympiad · Estonia

Today, on September 23, 2023, twins Mari and Jüri received a total of $5000$ candies for their $10$ th birthdays. Starting from this day, their mother allows them both to take candies once per day, such that the amount of candies taken by any child on any day is less than their age in full years (on their birthday, they already use their new age). Neither child can resist taking at least one candy every day. Jüri allows Mari to take candies first on every day. The children agreed that whoever takes the last candy, has to buy new candies. Which child can ensure that they do not have to buy new candies, no matter how the other child takes their candies?

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

We will show that Mari can avoid taking the last candy. To achieve this, she will take $8$ candies today, and on every following day, she will take candies so that along with Jüri's candies from the previous day, the total is the age of the children on the previous day. Since $2024$ is a leap year, they will be $10$ years old for $366$ days; thus on the $11$ th birthday of the children, there will be $5000 - 8 - 366 \cdot 10 = 1332$ candies remaining. After $121$ more days, there will be $1332 - 121 \cdot 11 = 1$ candy remaining after Mari has taken her candies for the day. Thus Jüri has no choice but to take the last candy.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.