Competition problem — Number theory

Library / /55 of 196

Number theory Difficulty 4.7 AIME Prove it Soviet Union

Problem:

Bus numbers have 6 digits, and leading zeros are allowed. A number is considered lucky if the sum of the first three digits equals the sum of the last three digits. Prove that the sum of all lucky numbers is divisible by 13.

Solution

Solution:

The total is made up of numbers of the form $abcabc$ , and pairs of numbers $abcxyz$ , $xyzabc$ . The former is $abc \times 1001$ and the sum of the pair is $1001(abc + xyz)$ . So the total is divisible by $1001$ and hence by $13$ .

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