Competition problem — Number theory

Library / /15 of 28

Number theory Difficulty 5.2 AIME, harder Prove it Ukraine

Find all natural numbers $N$ , that have two digits and are equal to the sum of their digits added to the cube of this sum.

Solution

Let $n$ be the sum of the digits, then: $N = n + n^3$ . The cube has to be less than $99$ , that gives us four options: $1$ , $2$ , $3$ , $4$ . An easy check shows that the only possible answer is $30$ .

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.