Show that $\displaystyle 98^{76543} + 21$ is not divisible by $\displaystyle 11$.
I do not know how to approach this.
Since $\displaystyle n = 76543$ is odd, so $\displaystyle 98^{76543} - 1 \equiv 9 (mod 11)$. We also have $\displaystyle 20 \equiv 9 (mod 11)$. Hence $\displaystyle 98^{76543} - 1 - 20 \equiv 9 - 9 (mod 11)$ or $\displaystyle 98^{76543} - 21 \equiv 0 (mod 11)$. Doesn't this mean that $\displaystyle 98^{76543} - 21$ is divisible by $\displaystyle 11$?
How would you work out the congruence relations you stated by yourself say in exam conditions?
Thanks! Sometimes I get really confused.
So $\displaystyle 22 \equiv 0 (\mod 11)$ and add this to your congruence we get $\displaystyle 98^{76543} - 1 + 22 \equiv 9 + 0 (\mod 11)$ which is $\displaystyle 98^{76543} + 21 \equiv 9 (\mod 11)$.
But still, how do you know that $\displaystyle 98^{n} -1 \equiv 9 (\mod 11)$ when $\displaystyle n$ is odd?
Thanks, this is brilliant. Did you observer that $\displaystyle 98 \equiv -1 (\mod 11)$ and then work on this congruence like
$\displaystyle 98 \equiv -1 \pmod{11}$
$\displaystyle 98^{76543} \equiv (-1)^{76543} \pmod{11}$
$\displaystyle 98^{76543} \equiv -1 \pmod{11}$
$\displaystyle 98^{76543} + 21 \equiv -1 + 21 \pmod{11}$
$\displaystyle 98^{76543} + 21 \equiv 20 \pmod{11}$
$\displaystyle 98^{76543} + 21 \equiv 9 \pmod{11}$?
Well, I didn't do it in exactly that order, but that's equivalent. The steps can be written for example like this:
$\displaystyle 98^{76543} \equiv (-1)^{76543} \equiv -1 \pmod{11}$
$\displaystyle 21 \equiv 10 \pmod{11}$
$\displaystyle 98^{76543}+21 \equiv -1 + 10 \equiv 9 \pmod{11}$
But anyone reasonably familiar with congruences should be able to follow the one-liner I wrote above without any intermediate steps.
In general, in the context of an exam, you may be presented with $\displaystyle a^b \pmod{n}$ such that gcd(a,n)=1 and be expected to recognise that Euler's theorem applies. But if $\displaystyle a$ is congruent to 1, 0, or -1 (mod n), then you should definitely take advantage of that to make calculations very easy. And there are other related topics you can learn about if you are interested in such things.
A little of both, I guess. The Chinese Remainder Theorem can be used if gcd(a,n)>1, and (mainly discussed in a computer science context) there's an algorithm for fast modular exponentiation by repeated squaring. Using such concepts you could compute things like the common residue of $\displaystyle a^{b^{c^d}} \pmod{n}$ for nontrivial a,b,c,d,n. (And the numbers can get very large very quickly; for example, $\displaystyle 2^{2^{3^3}}$ has 40403563 digits.) You probably won't get something like that on a test, but nevertheless it's possible that having extra knowledge might provide a shortcut for certain problems.