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All-Soviet-Union Mathematical Olympiad

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 $13$ .

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