Discussion
What is the remainder when \(2^{31}\) is divided by 5 ?
Answer: C
Correct answer: 3
The powers of 2 modulo 5 repeat in a cycle of four.
\[2^1 \equiv 2, \quad 2^2 \equiv 4, \quad 2^3 \equiv 3, \quad 2^4 \equiv 1 \pmod 5\]
Since \(31 = 4 \times 7 + 3\), the exponent falls on the third entry of the cycle, so the remainder is 3.
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