Maths Olympiad Prep

Track / Stage 3 / 200 of 260 #200 of 1964

Problem 200

AMC 10/12, early questions
Number theory Difficulty 3.7 Find the answer South African Mathematics Olympiad Second Round · South Africa

What is the remainder when 220162^{2016} is divided by 1313?

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

(We do this by inspection, by trying out some values until we can see the pattern.) Draw up a list of the remainders left by the powers of 22 after division by 1313:

nn0123456789101112
2nmod132^n \bmod 131248361211951071

We see that the remainders repeat every 1212 terms, since 2121(mod13)2^{12} \equiv 1 \pmod{13}.

Now, 2016÷12=1682016 \div 12 = 168 with remainder 00, so 2016=12×168+02016 = 12 \times 168 + 0.

Therefore,
22016201(mod13). 2^{2016} \equiv 2^0 \equiv 1 \pmod{13}.

So the remainder is 11.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.