Question

Find the remainder when ${128}^{129}$ is divided by 17.

• A. 8
• B. 9
• C. 17
• D. 4

Answer 1

The correct option is B 9

Since the number 128 is not divisible by the prime 17, so we get
${128}^{16}$$\equiv$ 1 (mod 17) (1)
by Fermat’s Little Theorem. By the division algorithm,
129 = 16 $\times$ 8 + 1
Since ${128}^{129}={128}^{(16\times 8+1)}=({128}^8{)}^{16}\times {128}^1$, then
${128}^{129}\equiv 1^8\times 128\equiv 9$ (mod 17). [using (1)]
Hence, 9 is the answer.