Problem B Show that 3 · 4n + 51 is divisible by 3 and 9 for all positive integers n.

For any integer n ≥ 1, let $P_n$ be the statement that $3\cdot4^n+51$ is divisible by 9.

Base case. The statement $P_1$ says that $3\cdot4^1+51=63$ is divisible by 9, which is true.

Inductive Step. Fix k ≥ 1, and suppose that $P_k$ holds, that is, $3\cdot4^k+51$ is divisible by 9,

$3\cdot4^k+51=9m, m\in \Z, m\geq7$

It remains to show that $P_{k+1}$ holds, that is, $3\cdot4^{k+1}+51$ is divisible by 9,

$3\cdot4^{k+1}+51=4\cdot(3\cdot4^k)+51=$$=4\cdot(3\cdot4^k+51)-4\cdot51+51=$

$=4\cdot9m-3\cdot51=9(4m-17),m\in \Z, m\geq7$

Therefore $P_{k+1}$ holds.

Thus, by the principle of mathematical induction, for all n ≥ 1, $P_n$ holds.

Therefore $3\cdot4^n+51$ is divisible by 9 for all positive integers n.

Therefore $3\cdot4^n+51$ is divisible by 3 for all positive integers n.