Which of the following statements are correct?
(A) \(2^{3} \equiv 1 \pmod{7}\)
(B) \(2^{30} \equiv 2 \pmod{7}\)
(C) \(2^{31} \equiv 4 \pmod{7}\)
(D) \(2^{32} \equiv 4 \pmod{7}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) and (D) only
Since \(2^{3} = 8 = 7 + 1\), the powers of 2 repeat in a cycle of 3 modulo 7.
\(2^{3} \equiv 1 \pmod{7}\)
\(2^{30} = (2^{3})^{10} \equiv 1^{10} = 1 \pmod{7}\)
\(2^{31} = 2^{30} \times 2 \equiv 2 \pmod{7}\)
\(2^{32} = 2^{30} \times 4 \equiv 4 \pmod{7}\)
So \(2^{3}\), \(2^{30}\), \(2^{31}\) and \(2^{32}\) leave remainders 1, 1, 2 and 4 on division by 7.
Checking the statements
(A) \(2^{3} \equiv 1 \pmod{7}\) — Correct. \(8 - 1 = 7\).
(B) \(2^{30} \equiv 2 \pmod{7}\) — Incorrect. 30 is a multiple of 3, so the remainder is 1.
(C) \(2^{31} \equiv 4 \pmod{7}\) — Incorrect. \(2^{31} \equiv 1 \times 2 = 2\).
(D) \(2^{32} \equiv 4 \pmod{7}\) — Correct. \(2^{32} \equiv 1 \times 4 = 4\).