Question:
The remainder when \(7^{89}\) is divided by 15 is
The remainder when \(7^{89}\) is divided by 15 is
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Find the cycle of powers of 7 modulo 15 (or write 49 = 45 + 4).
Updated On: Nov 12, 2026
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The Correct Option is B
Solution and Explanation
\(7^2 = 49 = 3 \times 15 + 4 \equiv 4 \pmod{15}\), so \(7^4 \equiv 16 \equiv 1 \pmod{15}\).
\(89 = 4 \times 22 + 1\), so \(7^{89} = (7^4)^{22} \cdot 7 \equiv 1 \cdot 7 = 7 \pmod{15}\).
The remainder is 7.
Answer: (2) 7
\(89 = 4 \times 22 + 1\), so \(7^{89} = (7^4)^{22} \cdot 7 \equiv 1 \cdot 7 = 7 \pmod{15}\).
The remainder is 7.
Answer: (2) 7
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