# How do I compute $a^b\,\bmod c$ by hand?

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 $(a,c) \neq 1$, for instance $6^{103}\,\bmod 14$?

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

This is being asked in an effort to cut down on duplicates, see here: Coping with *abstract* duplicate questions.

• fermat's little theorem – geraldgreen Nov 11 '11 at 22:37
• It may be laborious when $c$ is huge, but $b$ being huge should be no problem. – Michael Hardy Nov 11 '11 at 22:45

## 9 Answers

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, $\phi(21) = \phi(3) \times \phi(7) = 2 \times 6 = 12$. $844325 \bmod 12 = 5$, so $5^5 = 5 \times 25^2 \equiv 5 \times 4^2 = 80 \equiv 17 \mod 21$.

• When $a$ and $c$ are coprime, but $0<b<\phi(c)$, 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. – user26486 Feb 18 '15 at 9:56
• How did you arrive at $$a^b \equiv a^{b \, \bmod \, \phi(c)} \, \bmod c$$ using Euler's theorem? – Victor Jun 9 '17 at 20:41
• By Euler theorem power of $a$ to Euler totient of $c$ is one module $c$, hence the exponent $b$ can be reduced modulo the totient. – Sasha Jun 9 '17 at 20:48
• $$a^b\equiv a^{\phi(c)x+y}\equiv (a^{\phi(c)})^xa^y\equiv a^y\bmod c$$ – Jaycob Coleman Apr 16 at 3:03

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. – Michael Hardy Nov 11 '11 at 23:05
• This method won't be so nice when $(5,21)\neq 1$... – mathmath8128 Nov 11 '11 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 $6^{844325} \equiv 6 \mod 21$. – Henry Nov 12 '11 at 1:38
• Beautiful! $(+1)$ – Mr Pie May 22 '18 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. – Rosie F Mar 18 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. – Jyrki Lahtonen Jun 9 '16 at 9:23
• If memory serves, this is sometimes referred to as the "Russian peasant" method. – J. M. is a poor mathematician Jun 9 '16 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.) – Rosie F May 28 '18 at 17:48

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 $16^{74}=(2^4)^{74}=2^{296} = 2^{6\cdot49}\cdot2^2 \equiv (-1)^{49}\cdot4 \equiv -1 \pmod{65}$. 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 $a^{p-1}\equiv1\pmod p$.)

• 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 $O(\log (n))$.

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 $n$.

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 it's place (another name for a)
• if a>c, take a mod c first.
• etc.

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

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