Prove that $1835^{1910} + 1986^{2061}$ is divisible by 7. So far I have that $1835=262\times7+1$ and $1986=283\times 7+5$. Then I have $1^{1910}+5^{2061} \mod 7$. I am not sure how to implement Fermat's little theorem to finish this. I know the theorem says if p is a prime and n is a positive integer then $p|n^p-n$. I googled the problem and found some answers but none of them said why exactly we use flt. http://www.thomasbeatty.com/MATH%20PAGES/ARCHIVES%20-%20EXAMS/Number%20Theory/10%20S%20Num%20Theory%20Test%202B%20-%20solutions.pdf
Prove that $1835^{1910} + 1986^{2061}$ is divisible by 7.
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number-theory
modular-arithmetic
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0Why do you need to use FLT? Once you know that $1835\equiv 1986 \equiv 5 \pmod 7$ then it just become simple computing of $5^n \pmod 7$ for $n\ge 1$. This shoud be a cycle of $5\to 4 \to 6\to 2\to 3\to 1\to 5$ – 2017-02-19
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0Hint: $2061=6*343+3$ – 2017-02-19
2 Answers
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Lil' Fermat says that if $n\not\equiv 0\mod 7$, $n^6\equiv 1\mod 7$, so that $n^k\equiv n^{k\bmod 6}\mod 7$.
So, here, $$1835^{1910}+1986^{2061}\equiv 1+5^3\equiv 1+(-2)^3\equiv 1-8\equiv1-1=0\mod 7.$$
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1I just about understand why US mathematicians find it cute to talk about "Baby Rudin". But taken together with "Lil' Fermat" the vision of a Disney movie full of odious mathematical prodigies in their infancy is just too puke-making. $\ddot{\smile}$. – 2017-02-19
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0Very smart! Actually, in my case , it refers to an old comic strip from the thirties, Lil' Abner. As to odious mathematical prodigies, I'd rather imagine some TeX Avery's cartoon… – 2017-02-20
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0@Rob It get's better: $\mu\rm Fermat$, and Gosper's milli-Ramanujan scale, etc. The sky is the limit... – 2017-02-20
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0@Bill Dubuque: you should propose it to the maintainer of the `siunitx` package ;o) – 2017-02-20
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$1^{1910} \equiv 1 \pmod 7$
$5^{2061} \equiv 5^3 \pmod 7$ since $(5^6)^{343} = 5^{2058} \equiv 1 \pmod 7$
$7 \mid 126$