Discussion

What is the remainder when \(4^{19}\) is divided by 33?

Answer: D

Correct answer: 25

Rather than expanding \(4^{19}\), find how soon the powers of 4 repeat modulo 33.

\[4^1 = 4,\quad 4^2 = 16,\quad 4^3 = 64 \equiv 31,\quad 4^4 \equiv 124 \equiv 25,\quad 4^5 \equiv 100 \equiv 1 \pmod{33}\]

The cycle length is 5. Since \(19 = 5\times 3 + 4\),

\[4^{19} = \left(4^5\right)^3 \times 4^4 \equiv 1^3 \times 25 = 25 \pmod{33}\]

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