Discussion

What is the remainder if \(13^{73} + 14^{3}\) is divided by 11?

Answer: D

Correct answer: 2

Reduce each term modulo 11 separately.

Since \(13 \equiv 2 \pmod{11}\) and Fermat’s little theorem gives \(2^{10} \equiv 1 \pmod{11}\), write \(73 = 10\times 7 + 3\):

\[13^{73} \equiv 2^{73} = \left(2^{10}\right)^{7}\times 2^{3} \equiv 1\times 8 = 8 \pmod{11}\]

And \(14 \equiv 3\), so \(14^{3} \equiv 3^{3} = 27 \equiv 5 \pmod{11}\).

Adding the two parts, \(8 + 5 = 13 \equiv 2 \pmod{11}\).

No comment is present. Be the first to comment.

Loading…