Find the smallest positive integer n such that 3^n ≡ 1 (mod 2^10).
Correct: A
We need the multiplicative order of 3 modulo 1024. Since 3 is odd, it is invertible. The group of units mod 1024 has order φ(1024)=512. By the lifting-the-exponent lemma for orders, ord_{2^k}(3)=2^{k-2} for k≥3. Thus ord_{1024}(3)=2^{10-2}=256.