Discussion
Find the remainder when \(2^{50}\) is divided by 7.
Answer: C
We can find the cyclicity of remainders for powers of 2 mod 7. \(2^1\equiv 2\), \(2^2\equiv 4\), \(2^3=8\equiv 1\). The cycle length is 3. We need to find the remainder of the power \(50\) when divided by 3. \(50 \div 3\) gives a remainder of 2. So the required remainder is the 2nd in the cycle (2, 4, 1), which is 4. Alternatively, using Fermat's Little Theorem, since 7 is prime, \(2^6 \equiv 1 \pmod{7}\). Then \(2^{50} = (2^6)^8 \times 2^2 \equiv 1^8 \times 4 \equiv 4 \pmod{7}\).
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