Krdytkyu

Let's try $5^{844325} bmod 21$:
$$
begin{align}
5^0 & & & equiv 1 \
5^1 & & &equiv 5 \
5^2 & equiv 25 & & equiv 4 \
5^3 & equiv 4cdot 5 & & equiv 20 \
5^4 & equiv 20cdot 5 & & equiv 16 \
5^5 & equiv 16cdot 5 & & equiv 17 \
5^6 & equiv 17cdot 5 & & equiv 1
end{align}
$$
So multiplying by $5$ six times is the same as multiplying by $1$. We want to multiply by $5$ a large number of times: $844325$. How many times do we multiply by $5$ six times? The number of times $6$ goes into $844325$ is $140720$ with a remainder of $5$. That remainder is what matters. Multiply by $5^6$ exactly $140720$ times and that's the same as multiplying by $1$ that many times. Then multiply by $5$ just $5$ more times, and get $17$.

So $5^{844325} equiv 17 bmod 21$.

Here are two examples of the square and multiply method for $5^{69} bmod 101$:

$$ begin{matrix}
5^{69} &equiv& 5 &cdot &(5^{34})^2 &equiv & 37
\ 5^{34} &equiv& &&(5^{17})^2 &equiv& 88 &(equiv -13)
\ 5^{17} &equiv& 5 &cdot &(5^8)^2 &equiv& 54
\ 5^{8} &equiv& &&(5^4)^2 &equiv& 58
\ 5^{4} &equiv& &&(5^2)^2 &equiv& 19
\ 5^{2} &equiv& &&(5^1)^2 &equiv& 25
\ 5^{1} &equiv& 5 &cdot &(1)^2 &equiv& 5
end{matrix} $$

The computation proceeds by starting with $5^{69}$ and then working downward to create the first two columns, then computing the results from the bottom up. (normally you'd skip the last line; I put it there to clarify the next paragraph)

As a shortcut, the binary representation of $69$ is $1000101_2$; reading the binary digits from left to right tell us the operations to do starting from the value $1$: $0$ says "square" and $1$ says "square and multiply by $5$".

The other way is to compute a list of repeated squares:

$$ begin{matrix}
5^1 &equiv& 5
\ 5^2 &equiv& 25
\ 5^4 &equiv& 19
\ 5^8 &equiv& 58
\ 5^{16} &equiv& 31
\ 5^{32} &equiv& 52
\ 5^{64} &equiv& 78
end{matrix} $$

Then work out which terms you need to multiply together:

$$ 5^{69} equiv 5^{64 + 4 + 1} equiv 78 cdot 19 cdot 5 equiv 37 $$

Some tricks which are useful for modular exponentiation

The intention of this post is to collect various tricks which can sometimes simplify computations of this type. (Especially when done by hand and not using computer or calculator.) This post is community-wiki, so feel free to edit it if you have some ideas for improvements.

Using complement: $(c-a) equiv (-a) pmod c$

If the given number is close to $c$ (but smaller than $c$), replacing it by $c-a$ my help us - we will work with smaller numbers. Some examples:

• If we want to calculate $7^{777} bmod 50$, it is useful to notice that $7^2=49 equiv (-1) pmod{50}$, so we can replace $7^2$ by $-1$
  and get $7^{777} equiv 7^{388} cdot 7 equiv (-1)^{388} cdot 7 equiv 7 pmod{50}$. (This was part of Find $3^{333} + 7^{777}pmod{ 50}$.)


• We want to calculate $50^{50} bmod 13$. Since $4cdot 13 = 52$, we have $50 equiv -2 pmod{13}$. So we can work with $-2$ instead of $50$, which will be easier, since it is a smaller number. How to use Fermat's little theorem to find $50^{50}pmod{13}$?
  

If you can find a power which is close to the modulo, try to use it

Some examples:

• We want to calculate $6^{1000} bmod 23$. Since $6=2cdot 3$, let us have a look whether we can somehow combine these two numbers to get something with small remainder modulo $23$. We may notice that $24=2^3cdot 3 equiv 1pmod{23}$. We can also notice that $27 equiv 4pmod{23}$, i.e. $3^3equiv 2^2pmod{23}$. Replacing $2^2$ with $3^3$ in the previous congruence we get $2cdot 3^4 equiv 1 pmod{23}$. Now we can combine the preceding two congruences to get $1equiv (2^3cdot 3)^3cdot(2cdot 3^4)^2 = 2^{11}cdot3^{11} = 6^{11}pmod{23}$. Notice that the congruence $6^{11}equiv1pmod{23}$ can be obtained also by different means: Find $6^{1000} mod 23$
  


• We want to find $5^{119} bmod 59$. This can be solved in a very simple way using Fermat's little theorem: Find the remainder using Fermat's little theorem when $5^{119}$ is divided by $59$? However, let us forget Fermat's little theorem and let us try to find some powers of $5$ which give small remainder modulo $59$. We may notice that $5^3$ is not too far from $2cdot59$ and get $5^3equiv125equiv7pmod{59}$. Similarly, $7cdot25$ seems to be not very far from $3cdot59$, so we can try $5^5=5^3cdot5^2equiv7cdot25equiv175equiv-2pmod{59}$. And now we can use that $64$ is a power of two which is close to our remainder to get $5^{30} = (5^5)^6 equiv (-2)^6 equiv 64 equiv 5 pmod{59}$. Since we have $5^{30}equiv5pmod{59}$ and $gcd(5,59)=1$, we can cancel $5$ on both sides to get $5^{29}equiv1pmod{59}$. And the last fact can be used in further computations.


• The task is to find $16^{74} bmod 65$. One may notice that $64$ is a power of two which is very close to $65$. So we have $2^6 = 64 equiv -1 pmod{65}$, meaning that $16^{74}=(2^4)^{74}=2^{296} = 2^{6cdot49}cdot2^2 equiv (-1)^{49}cdot4 equiv -1 pmod{65}$. See also https://math.stackexchange.com/questions/2077609/computing-1674-bmod-65.

Using Euler's criterion

Euler's criterion can tell us about value of $a^{frac{p-1}2}$ modulo a prime $p$. However, we need to know whether $a$ is a quadratic residue modulo $p$. For some numbers this can be guessed. Sometimes this can be checked using quadratic reciprocity (Of course, this is not much of an improvement in comparison with Fermat's little theorem, which gives us $a^{p-1}equiv1pmod p$.)

• Let us have a look at $5^{29} bmod 59$ (we have already computed this using different computations above). It is easy to notice that $8^2=64equiv5pmod{59}$, so $5$ is a quadratic residue modulo $59$. So from Euler's criterion we get $5^{29}=5^{(59-1)/2}equiv1pmod{29}$.

In general, squared exponentiation is used, this is $O(log(b) cdot log(n))$ if multiplication $bmod n$ is $O(log (n))$.

powmod(a,b,n){

    res=1;

    fact=a;

    while(b>0){

        res=(res*(b%2)==1?fact:1)%n;

        fact=(fact*fact)%n;

        b=floor(b/2);

    }

}

when $b$ is huge (much larger than $n$) you can (attempt) to find the rank of the ring ($varphi(n)$) and find the remainder of $b pmod {varphi(n)}$ because $a^b bmod n= a^{b mod varphi(n)} bmod n$ (for $21$, it is $(3-1) cdot (7-1)=12$) this requires finding the prime factors of $n$.

In general the rank for $n = prod{(p_i)^{k_i-1} cdot (p_i-1)}$ with $p_i^{k_i}$ the prime factors of $n$.

The Chinese remainder theorem can reduce the computation needed. For example, we can factor $21 = 3 cdot 7$, and have

$$ 1 cdot 7 - 2 cdot 3 = 1$$

(in general, we can use the extended Euclidean algorithm to produce this formula)

Consequently, if

$$x equiv a pmod 3 qquad x equiv b pmod 7 $$

then

$$ x equiv a cdot (1 cdot 7 ) + b cdot (-2 cdot 3) pmod{21} $$

Thus, we can compute $5^{844325} bmod 21$ by using our favorite means to compute:

$$ 5^{844325} equiv 2 pmod 3 qquad 5^{844325} equiv 3 pmod 7 $$

and thus

$$ 5^{844325} equiv 2 cdot 7 + 3 cdot (-6) equiv -4 equiv 17 pmod{21} $$

For the first question: use $a^{Phi(c)}=1 mod c$, where $Phi(c)$ is the number of coprimes to $c$ below $c$. For $c=21=7cdot 3$ we have $Phi(c)=(7-1)cdot(3-1)=12$

second question: Use $a^4=(a^2)^2, a^8=(a^4)^2$ and so on. Decompose the exponent into powers of 2 and combine them using $a^ncdot a^m=a^{n+m}$ E.g. $a^{69}=a^{64}cdot a^4cdot a^1$

Its not hard to show that the sequence

$$
x_n=a^nmod{c}
$$

is periodic, with period $p$ (which is at most $c$). Evaluate the first few terms to get the period ${x_0,x_1,dots,x_{p-1}}$. Then you can evaluate for any huge power $n$ as

$$
x_n=x_{nmod{p}}
$$