Discussion

What is the remainder when \(4^{21}\) is divided by 33?

Answer: B

We can use Euler's Totient Theorem. The totient of 33, \(\phi(33) = \phi(3 \times 11) = \phi(3) \times \phi(11) = (3-1)(11-1) = 2 \times 10 = 20\).

According to Euler's theorem, if a and n are coprime, then \(a^{\phi(n)} \equiv 1 \pmod{n}\). Here, 4 and 33 are coprime.

So, \(4^{20} \equiv 1 \pmod{33}\).

We need to find the remainder of \(4^{21} \div 33\).

\(4^{21} = 4^{20} \times 4^1\).

So, \(4^{21} \pmod{33} \equiv (4^{20} \pmod{33}) \times (4^1 \pmod{33}) \pmod{33}\).

\(\equiv 1 \times 4 \pmod{33}\).

The remainder is 4.

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