Discussion

What is the remainder when \(3^{10}\) is divided by 7?

Answer: C

We can use Fermat's Little Theorem. Since 7 is a prime number, \(3^{7-1} = 3^6 \equiv 1 \pmod{7}\). We need to find the remainder for \(3^{10}\). We can write \(3^{10} = 3^6 \times 3^4 \equiv 1 \times 3^4 \pmod{7}\). Now we just need to calculate \(3^4 = 81\). The remainder of \(81 \div 7\) is 4 (since \(81 = 11 \times 7 + 4\)). So the remainder is 4.