Maths Olympiad Prep

Track / Stage 6 / 12 of 400 #1012 of 1964

Problem 1012

National Olympiad, first round
Algebra Difficulty 6.0 Prove it Ukrainian National Mathematical Olympiad · Ukraine

Andriy and Olesya write a natural number each on a chalkboard. It turns out, that number, written by Olesya, has sum of digits 20182018 and has precisely 11 digit less than Andriy's number. It is also known, that difference of numbers, written by him, equals to one-digit number. What can be the number, written by Andriy?

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.

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Official solution

It is not hard to see, that Andriy's number can only be 100...0a\overline{100...0a}, and Olesya's – only: 99...9b\overline{99...9b}. Otherwise, the difference will not be a one-digit number. Really, if not all Olesya's digits, except for the last, are 99, then after adding a one-digit number, the number of digits will not change. So this is the presentation of Andriy's number. The sum of digits of Olesya's number equals 20182018, so it is 99...92\overline{99...92} (as 2018=2249+22018 = 224 \cdot 9 + 2). So Andriy's number has to have the last digit less than 22, because otherwise the difference of written numbers will not be less than 1010. So, this number can be 00 or 11. So, he wrote number 100...0\overline{100...0}, or 100...01\overline{100...01}.

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