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Question 1: Is $21^{100}-12^{100}$ divisible by $11$?

My work: Note:$$\begin{align*} & \color{red}{21^{100}}\equiv10^{100}\equiv 100^{50}\equiv1^{50}\equiv1 \mod 11\\ & \color{blue}{12^{100}}\equiv 1^{100}\equiv 1\mod 11\end{align*}$$ Hence,$$\begin{align*}\color{red}{21^{100}}-\color{blue}{12^{100}}\equiv 1-1\equiv 0\mod 11\end{align*}$$ Hence, $21^{100}-12^{100}$ is divisible by $11$.

Actual Question:

  1. Does my work hold? Is it logical and reasonable?
  2. Is there a quicker way of solving this problem?

I am very new to modular arithematic (i.e I just learned it $10$ minutes ago) so tips on a simple book where I can learn a bit more in-depth stuff about modular arithematics greatly helps!

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    Working modulo $11$ is correct, logical and reasonable. It is a bit shorter if you reduce $21$ modulo $11$ to $-1$. – 2017-01-26
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    Yes, it looks correct. A slightly quicker way for the first part is noting that $10 \equiv -1$ and $(-1)^{100} \equiv 1$ because $100$ is even. – 2017-01-26
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    It's fine but relies a little on lucky calculations. Provided $a,b$ are both coprime to $11$, $a^{10k}- b^{10k}$ is always divisble by $11$. Check [Fermat's little theorem](https://en.wikipedia.org/wiki/Fermat's_little_theorem) – 2017-01-26
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    @Joffan: I'd say it is completely reasonable to check whether the base is trivial before trying to reduce the exponent, in cases such as this. – 2017-01-26
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    @HenningMakholm agreed, and if my comment appeared to be critical of the OP's work, I apologize - that wasn't my intent. – 2017-01-26

4 Answers 4

0

$21^{100}-12^{100}\equiv21^0-12^0\equiv0($mod $11)$

since

$21^{2k}\equiv12^{k}\equiv1($mod $11)$

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    Your first step there is not at all well-expressed. I could see that misleading people. – 2017-01-26
2

An easier approach for such problems is making use of Fermat's_little_theorem which says, for any integer $a$,
$a^{p-1}\equiv 1 \mod{p}$
if $p$ is prime and $gcd(a,p)=1$

Using this, we can solve the problem as follows.

$21^{11-1}\equiv 1 \pmod {11}\\ 21^{10}\equiv 1 \pmod {11}\\ 21^{100}\equiv 1 \pmod {11} \cdots(1)$

$12^{11-1}\equiv 1 \pmod {11}\\ 12^{10}\equiv 1 \pmod {11}\\ 12^{100}\equiv 1 \pmod {11} \cdots(2)$

Therefore $21^{100}-12^{100} \equiv 0 \pmod {11}$

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    thanks. you are right. i have added that into answer – 2017-01-26
2

Your calculation looks correct. It would be more conventional (and slightly less work) to write it as: $$ 21\equiv -1 \pmod{11} \qquad\text{and}\qquad 12\equiv 1 \pmod{11} $$ so $$ 21^{100} - 12^{100} \equiv (-1)^{100} - 1^{100} \equiv 1 - 1 \equiv 0 \pmod{11} $$ since $-1$ to an even power is always $1$.

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    So how would you calculate $2013^3\equiv ?\mod 2015$ using the negative remainder thing? – 2017-01-26
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    $2013\equiv -2\ $ so .... – 2017-01-26
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    @Crescendo: $2013\equiv -2$, and $(-2)^3=-8$, so $2013^3\equiv 2007\pmod{2015}$. – 2017-01-26
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    @HenningMakholm Ah thanks! – 2017-01-26
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    An alternative approach is to go via $21\equiv -12$ – 2017-01-26
0

$$\begin{align} 21^{100} - 12^{100} & \equiv (11 \times 2 - 1)^{100} - (11 + 1)^{100} \\ & \equiv (0 \times 2 - 1)^{100} - (0 + 1)^{100} \\ & \equiv (-1)^{100} - 1^{100} \\ & \equiv 1 - 1 \\ & \equiv 0 \pmod {11} \end{align}$$

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    The point of modular arithmetic is to *ignore* obfuscatory quotients - not to highlight them as above. – 2017-01-26