1972 AMC 12 Problem 31

Attempt Problem 31 of the 1972 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1972 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

31.

When the number

2^{1000}

is divided by

13,

the remainder in the division is:

1

2

3

7

11

Use

2^6\equiv-1\pmod{13}

Write

1000=6\cdot166+4

Since

2^6=64\equiv-1\pmod{13},

\begin{aligned} 2^{1000}&=2^{6\cdot166+4}\\ &\equiv(-1)^{166}2^4\\ &\equiv16\\ &\equiv3\pmod{13}. \end{aligned}

Therefore, the correct answer is C.

Problem 31 in other years