Wednesday, Jan 11th

Problem
A 3 digit natural number is called tricubic if its the sum of the cubes of its digits. Find all pairs of consecutive tricubic numbers.

Solution
Suppose that $(n,n+1)$ is a pair of consecutive tricubic numbers and let $n=a10^2+b10+c$ be the decimal expansion of $n$. Then $n=a^3+b^3+c^3$ and $n+1=a^3+b^3+(c+1)^3$ (if $c\neq 9$) or $n+1=a^3+(b+1)^3$ (if $c=9$ and $b\neq 9$) or $n+1=(a+1)^3$ (if $c=b=9$).

In the first case, we have that $1=3c^2+3c+1$, from which $c=0$. Looking at the possibilities for $a$ and $b$ we find that there are no such tricubic numbers.

In the second case, we have that $9^3=3b(b+1)$, which has no solutions for $b$. In the third case, we have that $9^3+9^3=3a(a+1)$ which has no solutions for $a$.

Hence, there are no such pair of numbers.