# Modular exponentiation by hand ($a^b\bmod c$)

How do I efficiently compute $$a^b\bmod c$$:

• When $$b$$ is huge, for instance $$5^{844325}\bmod 21$$?
• When $$b$$ is less than $$c$$ but it would still be a lot of work to multiply $$a$$ by itself $$b$$ times, for instance $$5^{69}\bmod 101$$?
• When $$(a,c)\ne1$$, for instance $$6^{103}\bmod 14$$?

Are there any other tricks for evaluating exponents in modular arithmetic?

• See mod order reduction for the standard methods (with proofs) Mar 23 at 19:30

Wikipage on modular arithmetic is not bad.

• When $$b$$ is huge, and $$a$$ and $$c$$ are coprime, Euler's theorem applies: $$a^b \equiv a^{b \, \bmod \, \phi(c)} \, \bmod c$$ For the example at hand, $$\phi(21) = \phi(3) \times \phi(7) = 2 \times 6 = 12$$. $$\Rightarrow 844325 \bmod 12 = 5,\ \text{so}\ 5^5 = 5 \times 25^2 \equiv 5 \times 4^2 = 80 \equiv 17 \mod 21$$.

• When $$a$$ and $$c$$ are coprime, but $$0, repeated squaring (or using other compositions of powers) is the fastest way to go (manually): $$\begin{eqnarray} 5^4 \equiv 5 \times 5^3 \equiv 5 \times 24 \equiv 19 &\pmod{101}\\ 19^4 \equiv (19^2)^2 \equiv 58^2 \equiv (-43)^2 \equiv 1849 \equiv 31 &\pmod{101} \\ 31^4 \equiv (31^2)^2 \equiv (961)^2 \equiv 52^2 \equiv 2704 \equiv 78 &\pmod{101} \\ 5^{69} \equiv 5 \times 5^4 \times ((5^4)^4)^4 \equiv 5 \times 19 \times 78 \equiv 5 \times 19 \times (-23)\\ \equiv 19 \times (-14) \equiv -266 \equiv 37 & \pmod{101} \end{eqnarray}$$

• When $$a$$ and $$c$$ are not coprime, let $$g = \gcd(a,c)$$. Let $$a = g \times d$$ and $$c = g \times f$$, then, assuming $$b > 1$$: $$a^b \bmod c = g^b \times d^b \bmod (g \times f) = ( g \times (g^{b-1} d^b \bmod f) ) \bmod c$$ In the example given, $$\gcd(6,14) = 2$$. So $$2^{102} \times 3^{103} \mod 7$$, using Euler'r theorem, with $$\phi(7) = 6$$, and $$102 \equiv 0 \mod 6$$, $$2^{102} \times 3^{103} \equiv 3 \mod 7$$, so $$6^{103} \equiv (2 \times 3) \equiv 6 \mod 14$$.

• Carmichael function often reduces the exponent more than Euler's totient function. Feb 18, 2015 at 9:56
• When $\gcd(a,c)\ne1$, one can summarize the last case as$$a^b\equiv a^{(b\bmod\varphi)+\varphi}\pmod c$$where $\varphi=\varphi(c/\gcd(a^{\lfloor\log_2(c)\rfloor},c))$ pulls all common factors out at once and then applies Euler's totient theorem. As $\log_2(c)$ is usually very small, and the Euclidean algorithm let's us to use modular exponentiation (by squaring), $a^{\lfloor\log_2(c)\rfloor}$ is not a difficult computation. This has the advantage that we only require computing the $\gcd$ once and avoids repeatedly nesting mod's. Mar 17, 2020 at 19:31

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 4\cdot 5 & & \equiv 20 \\ 5^4 & \equiv 20\cdot 5 & & \equiv 16 \\ 5^5 & \equiv 16\cdot 5 & & \equiv 17 \\ 5^6 & \equiv 17\cdot 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 $5^{844325} \equiv 17 \bmod 21$.

• You may want to check my arithmetic, but this method will do it when you get that right. Nov 11, 2011 at 23:05
• This method won't be so nice when $(5,21)\neq 1$... Nov 11, 2011 at 23:53
• @aengle: it will be similar: for example for $6^{844325} \mod 21$ you would look at $6^0 \equiv 1$, $6^1 \equiv 6$, $6^2 \equiv 15$, $6^3 \equiv 6$, $6^4 \equiv 15$, so with a period of $2$. The number of times $2$ goes into $844325$ is $422162$ with a remainder of $1$. So $6^{844324} \equiv 15 \mod 21$ and thus $6^{844325} \equiv 6 \mod 21$. Nov 12, 2011 at 1:38
• Beautiful! $(+1)$ May 22, 2018 at 11:56
• Your search for 5's order mod 21 can be shortened: having found $5^3=-1\mod 21$ you may infer $5^6=1\mod 21$ immediately. You were lucky in this example that the order was small. This process takes time proportional to that order. Mar 18, 2019 at 21:04

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:

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

• Good to have an example of square-and-multiply at hand. Jun 9, 2016 at 9:23
• If memory serves, this is sometimes referred to as the "Russian peasant" method. Jun 9, 2016 at 10:20
• @J.M.isnotamathematician It is closer to $a$ to the power of the "Russian peasant" method. en.wikipedia.org/wiki/Ancient_Egyptian_multiplication (The named method is a multiplication algorithm using doubling and adding; what we have here is a power algorithm using squaring and multiplying.) May 28, 2018 at 17:48
• It is notable that what you've described isn't quite the usual exponentiation by squaring that is listed on Wikipedia and in this answer, which makes use of $a^{2n}=(a^2)^n$ instead of what you have, which is $a^{2n}=(a^n)^2$. This allows us to avoid having squares on the outside, which requires us to do more computations to fill in missing results. Dec 16, 2019 at 16:37

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 $$4\cdot 13 = 52$$, 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 $$6=2\cdot 3$$, 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 $$24=2^3\cdot 3 \equiv 1\pmod{23}$$. We can also notice that $$27 \equiv 4\pmod{23}$$, i.e. $$3^3\equiv 2^2\pmod{23}$$. Replacing $$2^2$$ with $$3^3$$ in the previous congruence we get $$2\cdot 3^4 \equiv 1 \pmod{23}$$. Now we can combine the preceding two congruences to get $$1\equiv (2^3\cdot 3)^3\cdot(2\cdot 3^4)^2 = 2^{11}\cdot3^{11} = 6^{11}\pmod{23}$$. Notice that the congruence $$6^{11}\equiv1\pmod{23}$$ 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 $$2\cdot59$$ and get $$5^3\equiv125\equiv7\pmod{59}$$. Similarly, $$7\cdot25$$ seems to be not very far from $$3\cdot59$$, so we can try $$5^5=5^3\cdot5^2\equiv7\cdot25\equiv175\equiv-2\pmod{59}$$. 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 $$5^{30}\equiv5\pmod{59}$$ and $$\gcd(5,59)=1$$, we can cancel $$5$$ on both sides to get $$5^{29}\equiv1\pmod{59}$$. 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 $$16^{74}=(2^4)^{74}=2^{296} = 2^{6\cdot49}\cdot2^2 \equiv (-1)^{49}\cdot4 \equiv -1\cdot 4 \equiv -4 \pmod{65}$$. See also Computing $$16^{74} \bmod 65$$.

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 $$a^{p-1}\equiv1\pmod p$$.)

• 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 $$8^2=64\equiv5\pmod{59}$$, so $$5$$ is a quadratic residue modulo $$59$$. So from Euler's criterion we get $$5^{29}=5^{(59-1)/2}\equiv1\pmod{29}$$.

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

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

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

def powmod(a, b, c):
res = 1
while b > 0:
if b % 2 == 1:
res = res * a % c
a = a * a % c
b //= 2
return res


Try it online

Example for $$5^{69}\bmod101$$:

\begin{align} 5^{69} & \equiv 5 \times (5^2)^{34} & \equiv 5 \times 25^{34} \\ & \equiv 5 \times (25^2)^{17} & \equiv 5 \times 19^{17} \\ & \equiv 5 \times 19 \times (19^2)^8 & \equiv 95 \times 58^8 \\ & \equiv 95 \times (58^2)^4 & \equiv 95 \times 31^4 \\ & \equiv 95 \times (31^2)^2 & \equiv 95 \times 52^2 \\ & \equiv 95 \times 78 \\ & \equiv 37 \end{align}

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 $$n$$.

For the first question: use $a^{\Phi(c)}=1 \mod c$, where $\Phi(c)$ is the number of coprimes to $c$ below $c$. For $c=21=7\cdot 3$ 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 $a^n\cdot a^m=a^{n+m}$ E.g. $a^{69}=a^{64}\cdot a^4\cdot a^1$

There are a few things of note:

• Exponent rules help. If b is a large composite, being the product of d,e,f,g,h,i,j,... then powering to b is like powering by d then e then f then g doing each in turn to your results, is easier (maybe as tedious) than one big computation.
• If a and c are coprime, then a raised to any power will also be coprime, so either you use up all coprime remainders or you don't but you can tell by powering until the remainder is 1, and 1 raised to any power is 1 letting you trim b down. (basically behind Euler and Fermat)
• if a and c are not coprime, then powers of a, sit at multiples of their gcd.
• Exponent rules help again if you find a sum equal to b you can use the product of same base powers = sum of exponents rule.(binary exponentiation uses this)
• if a is larger than half of c, use -(c-a) in its place (another name for a)
• if a>c, take a mod c first.
• etc.

Specifically in the case of $$\gcd(a,c)\ne1$$, we can use a generalization of Euler's totient theorem, which gives us:

$$a^b\equiv a^{(b\bmod\varphi)+\varphi}\pmod c$$

where $$b>\varphi=\varphi(c)$$.

Using the Chinese remainder theorem, this can be improved to $$\varphi=\varphi(c')$$, where $$c'$$ is the greatest factor of $$c$$ that is coprime to $$a$$. For a brute force computation of $$c'$$, one can use $$c'=c/\gcd(a^{\lfloor\log_2(c)\rfloor},c)$$.

When we have $$b<2\varphi$$, we can then apply exponentiation by squaring.

$$\varphi(c')=\varphi(7)=6$$, so $$\bmod14:$$

$$6^{103}\\\equiv6^{(103\bmod6)+6}\\=6^7\\=6\times36^3\\\equiv6\times8^3\\=48\times64\\\equiv6\times8\\=48\\\equiv6$$

Here we use a 'work out in place / lazy way / by hand' algorithm for the problem

$$\quad$$ Solve $$5^{69}\,\bmod 101$$.

$$\; 5^{69} = \big((4 + 1) 5^2\big)^{23} \equiv 24^{23}= 24 \big((4 + 20) {24}\big)^{11} \equiv 24\, (71^{11}) \equiv -24\, (30^{11}) =$$
$$\quad (-24)(30) \big((15 + 15) 30\big)^{5} \equiv (-24)(30)\, ({-9}^{5}) \equiv 24 \times 30 \times (-20) \times (-20) \times 9 \equiv$$ $$\quad 24 \times 30 \times (-4) \times 9 \equiv 24 \times (-19) \times 9 \equiv 24 \times (-70) \equiv 24 \times 31 \equiv$$
$$\quad (24 \times 4) \times 8 - 24 \equiv -64 \equiv 37 \,\bmod 101$$

Note: Since some discretion was used, we didn't actually specify an algorithm. But the work could be done to have a computer use simple lookup tables and produce similar outputs without using any math registers.

Adding an example for calculating the remainder of an iterated power.

Let's find the two last digits of $$97^{75^{63}}$$.

Equivalently, we want to find its remainder modulo $$100$$.

1. First we observe that $$\gcd(97,100)=1$$. If we had common prime factors here we would deal with each prime power separately using the Chinese remainder theorem. See also this answer (and the following three steps). Given that $$\phi(100)=40$$, we can immediately deduce that $$97^{40}\equiv1\pmod{100}$$.
2. Therefore we next need to determine the remainder of the exponent $$75^{63}$$ modulo $$40$$. Observe that $$\gcd(75,40)=5$$, so the power is obviously a multiple of five. We need to determine its residue class modulo $$40/5=8$$.
3. Modulo $$8$$ we have $$75\equiv3$$. Therefore $$75^{63}\equiv3^{63}\pmod 8$$. We see that $$3^2=9\equiv1\pmod8$$, so $$3^{63}\equiv3\pmod8$$.
4. So we know that $$75^{63}$$ is divisible by $$5$$ and leaves remainder $$3$$ modulo $$8$$. Because $$35$$ has these same remainders modulo $$5$$ and $$8$$, and $$\gcd(5,8)=1$$, the Chinese remainder theorem tells us that $$75^{63}\equiv35\pmod{40}.$$
5. The huge number $$97^{75^{63}}$$ is thus congruent to $$97^{35}\pmod {100}$$. Now we can either resort to exponentiation by squaring or use other tricks. Whatever we do, the end result is that $$97^{35}\equiv93\pmod{100},$$ so we can conclude that the two last digits are $$93$$.

Instead of the Euler totient function $$\phi(n)$$ you may consider using the Carmichael function $$\lambda(n)$$ instead. The workload may be reduced considerably. Particularly if an exponent has a small remainder modulo $$\lambda(n)$$, but a large remainder modulo $$\phi(n)$$.

Its not hard to show that the sequence

$$x_n=a^n\mod{c}$$

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

$$x_n=x_{n\mod{p}}$$

• Not only is this slow but you mod $n$ by the cycle length in that last equality first ... Aug 1, 2021 at 18:37