Competition problem — Number theory

Library / /45 of 520

Number theory Difficulty 4.9 AIME Find the answer

Problem 9.6. The teacher wrote a number on the board. Sasha decided to divide it by 102 with a remainder, and Masha - by 103. It turned out that the quotient obtained by Sasha and the remainder obtained by Masha add up to 20. What remainder did Sasha get? List all possible options.

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Answer: 20.

Solution. Let $n$ be the number on the board, divide it by 102 and by 103 with remainders:

$n=102 a+b ; \quad n=103 c+(20-a)$

Subtract the second equation from the first:

$0=103(a-c)+(b-20)$

From this, it follows that $b-20$ is divisible by 103. Since $0 \leqslant b \leqslant 101$ , we get that $b=20$ .

Notice also that, for example, for the number $n=122$ , the remainder $b=20$ is indeed possible.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.