Taken from a recent Mathletes event: What is the remainder of $212^{2017}$ divided by $213$? (Only the most basic TI calculator was allowed, so no brute force calculation)
What is the remainder of $212^{2017}$ divided by $213$?
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elementary-number-theory
modular-arithmetic
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0What do you know about arithmetic modulo $213$? (Or modulo some other number?) – 2017-01-15
2 Answers
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Hint: $212 \equiv -1 \pmod{213}$ and $(-1)^{2k+1} = -1$
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0OH, so $212$ has the same modulo as $-1$? So the modulo alternates between $1$ and $212$. Neat. $212$'s the final answer? – 2017-01-15
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0Yes. You're on the right track. modulo $n$ corresponds to the remainder you get after division by $n$. So modulo $213$ takes on values between "$0$" and $212$. On a side note, I would highly recommend you to practice some modular arithmetic. There's a lot of interesting practice material available on internet. Have a good day ahead. – 2017-01-15
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I think the "mod $213$" answer is best, but a more elementary observation is that
$$212^{2017} = (213-1)^{2017} = (\mbox{terms divisible by $213$ }) + 213^1(-1)^{2016} + (-1)^{2017}$$
by the binomial theorem. So the above equals $213*N + 213 -1$ which has remainder $212$ upon division by $213.$
A second, maybe more clever, observation:
$$212^{2017}+1 = 212^{2017}+1^{2017} = (212+1)(212^{2016} - 212^{2015}+ \cdots +1)$$
so we immediately see that $212^{2017}+1$ is a multiple of $213$, and the conclusion follows.