Modular Arithmetic Remainder Congruency such as $98765 ^ {54321} \bmod 7$

Question

I'm trying to figure out how to find the remainder of large numbers not in mod 10 by doing different examples.

Something that I'm working on is $98765 ^ {54321}$ (mod 7).

Can anyone help me find a general way to do these types of problems?

Answer 1

There are several methods you can apply to try to solve problems of the form:

$x \equiv n^a$ (mod $m$)

1. First, you can try to find the smallest positive $y \equiv n$ (mod $m$), and replace $n$ with $y$ in the original congruence. Note that it is sometimes helpful to replace it with $y-m$ instead of $y$ if $|y-m|<y$.

For example, if $x \equiv 11^a$ (mod 4), then you can replace the 11 with -1 or 3, as $11 \equiv 3 \equiv -1$ (mod 4).

In this case, it would probably be more helpful to use -1, as its absolute value is less than 3, and so it is easier to take powers of -1 than 3.

$x \equiv 11^a \equiv (3(4)-1)^a \equiv (3(0)-1)^a \equiv -1^a$ (mod 4)

So using this method, you now have:

$x \equiv y^a$ (mod $m$), where $y$ is a relatively small number.

If $a$ is large, and the number you are taking powers of is not 1 or -1, then there are 2 methods which are usually used to cancel the congruence down:

2. If $m$ is a prime, $p$, then you can use Fermat's little theorem to simplify the congruence:

   $a^p \equiv a$ (mod $p$)

   or more usefully:

   $a^{p-1} \equiv 1$ (mod $p$) if $a$ is not divisible by $p$.

https://en.wikipedia.org/wiki/Fermat%27s_little_theorem

For example, if $x \equiv 3^{1000}$ (mod 7), then using Fermat's little theorem, you know that: $3^6 \equiv 1$ (mod 7).

Then you can rewrite the congruence as: $x \equiv 3^{6(166) + 4}\equiv {(3^{6})}^{166} \times 3^4 \equiv {1}^{166} \times 3^4 \equiv 3^4$ (mod 7).

This can often allow you to simplify the congruence down to something you can calculate.

3. On the other hand, if Fermat's little theorem does not help (or does not help you simplify it down to small and manageable numbers), then try taking powers of $y$ (mod $m$), and seeing if these result in small numbers.

For example, if $x \equiv 5^{101}$ (mod 12), then you cannot use Fermat's little theorem as 6 is not prime.

However, notice that $5^2 \equiv 1$ (mod 12).

Hence: $x \equiv 5^{2(50) + 1} \equiv (5^2)^{50} \times 5^1 \equiv 1^{50} \times 5 \equiv 5$ (mod 12).

I hope this helps.

Answer 2

First reduce the base mod 7. Since $98765\equiv 2\mod 7$, we want to find $$2^{54321}\mod 7$$

Since 2 and 54321 are relatively prime, using Euler's theorem, we can reduce 54321 mod $\phi(7)$. Since $54321 \equiv 3 \mod 6$, we want to find $$2^3 \mod 7$$

The answer in this case is 1.