Past exam of the year 1 of the computer science course of the University of Oxford / 2017 / Discrete mathematics, continuous mathematics and probability / 2 / e

The base cases are $13=3(2)+7(1)$, $14=3(0)+7(2)$ and $15=3(5)+7(0)$. Suppose a counterexample exists and choose the least integer $n\ge 13$ failing the claim. It is not one of these base cases, so $n\ge 16$ and $n-3\ge 13$.

Minimality supplies nonnegative integers $x,y$ with $n-3=3x+7y$. Then $n=3(x+1)+7y$, a contradiction. This proof by minimal counterexample establishes the claim. Three consecutive base cases are needed because the reduction subtracts three.