To prove "$n^{3-n}$ is divisible by 6 for all n>=1": Base case: n=1, 0 is divisible by 6. Inductive step: assume $k^{3-k}$ is divisible by 6. Show (k+1)^3-(k+1) = k^{3+3k}^{2+3k+1-k-1} = ($k^{3-k}$)+3$k^{2+3k}$ = ($k^{3-k}$)+3k(k+1). First term divisible by 6 (hypothesis). Second term: k(k+1) is always even, so 3k(k+1) is divisible by 6. Sum divisible by 6. JEE tests this pattern with numbers like 7^n-1 is divisible by 6, or 4^n-3n-1 is divisible by 9.