Common induction divisibility problems: prove $n^{3-n}$ is divisible by 6, 4^n-3n-1 is divisible by 9, 7^n-1 is divisible by 6, x^{n-y}^n is divisible by x-y. Strategy: In the inductive step, express the (k+1) case using the k case plus a remainder that's independently divisible. For 4^(k+1)-3(k+1)-1 = 4*(4^k-3k-1)+9k: first term divisible by 9 (hypothesis, times 4), second term divisible by 9. The key algebraic trick is writing f(k+1) = constant*f(k) + (multiple of divisor). JEE tests these as verify/identify type MCQs.