= Solution
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 $n\geq13$ failing the claim. It is not one of these base cases, so $n\geq16$ and $n-3\geq13$.
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.
Back to article page
