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S(n) is sum of digits of n: n^3 = 8S(n)^3+6S(n)n+1

Source: RMO Hyderabad 2016 , P3 .

October 12, 2016
number theory

Problem Statement

For any natural number nn, expressed in base 1010, let S(n)S(n) denote the sum of all digits of nn. Find all positive integers nn such that n3=8S(n)3+6S(n)n+1n^3 = 8S(n)^3+6S(n)n+1.
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