Modular arithmetic/Introduction

Modular arithmetic is a special type of arithmetic that involves only integers.

The Basic Idea

Let's use a clock as an example, except let's replace the $\displaystyle 12$ at the top of the clock with a $\displaystyle 0$ . Starting at noon, the hour hand points in order to the following:

$\displaystyle 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, \ldots$

This is the way in which we count in modulo 12. When we add $\displaystyle 1$ to $\displaystyle 11$ , we arrive back at $\displaystyle 0$ . The same is true in any other modulus (modular arithmetic system). In modulo $\displaystyle 5$ , we count

$\displaystyle 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, \ldots$

We can also count backwards in modulo 5. Any time we subtract 1 from 0, we get 4. So, the integers from $\displaystyle -12$ to $\displaystyle 0$ , when written in modulo 5, are

$\displaystyle 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0,$

where $\displaystyle -12$ is the same as $\displaystyle 3$ in modulo 5. Because all integers can be expressed as $\displaystyle 0$ , $\displaystyle 1$ , $\displaystyle 2$ , $\displaystyle 3$ , or $\displaystyle 4$ in modulo 5, we give these give integers their own name: the modulo 5 residues. In general, for a natural number $\displaystyle n$ that is greater than 1, the modulo $\displaystyle n$ residues are the integers that are whole numbers less than $\displaystyle n$ :

$\displaystyle 0, 1, 2, \ldots, n-1.$

This just relates each integer to its remainder from the Division Theorem. While this may not seem all that useful at first, counting in this way can help us solve an enormous array of number theory problems much more easily!

Congruence

There is a mathematical way of saying that all of the integers are the same as one of the modulo 5 residues. For instance, we say that 7 and 2 are congruent modulo 5. We write this using the symbol $\displaystyle \equiv$ :

$\displaystyle 7 \equiv 2 \pmod{5}.$

The (mod 5) part just tells us that we are working with the integers modulo 5. In modulo 5, two integers are congruent when their difference is a multiple of 5. Thus each of the following integers is congruent modulo 5:

$\displaystyle -12 \equiv -7 \equiv -2 \equiv 3 \equiv 8 \equiv 13 \equiv 18 \equiv 23 \pmod{5}$

In general, two integers $\displaystyle a$ and $\displaystyle b$ are congruent modulo $\displaystyle n$ when $\displaystyle a - b$ is a multiple of $\displaystyle n$ . In other words, $\displaystyle a \equiv b \pmod{n}$ when $\displaystyle \frac{a-b}{n}$ is an integer. Otherwise, $\displaystyle a \not\equiv b \pmod{n}$ , which means that $\displaystyle a$ and $\displaystyle b$ are not congruent modulo $\displaystyle n$ .

Examples

$\displaystyle 31 \equiv 1 \pmod{10}$ because $\displaystyle 31 - 1 = 30$ is a multiple of $\displaystyle 10$

$\displaystyle 43 \equiv 22 \pmod{7}$ because $\displaystyle \frac{43 - 22}{7} = \frac{21}{7} = 3$ , which is an integer.

$\displaystyle 8 \not\equiv -8 \pmod{3}$ because $\displaystyle 8 - (-8) = 16$ , which is not a multiple of $\displaystyle 3$ .

$\displaystyle 91 \not\equiv 18 \pmod{6}$ because $\displaystyle \frac{91 - 18}{6} = \frac{73}{6}$ , which is not an integer.

Sample Problem

Find the modulo $\displaystyle 4$ residue of $\displaystyle 311$ .

Solution:

Since $\displaystyle 311 \div 4 = 77$ R $\displaystyle 3$ , we know that

$\displaystyle 311 \equiv 3 \pmod{4}$

and $\displaystyle 3$ is the modulo $\displaystyle 4$ residue of $\displaystyle 311$ .

Making Computation Easier

We don't always need to perform tedious computations to discover solutions to interesting problems. If all we need to know about are remainders when integers are divided by $\displaystyle n$ , then we can work directly with those remainders in modulo $\displaystyle n$ . This can be more easily understood with a few examples.

Addition

Problem

Suppose we want to find the units digit of the following sum:

$\displaystyle 2403 + 791 + 688 + 4339.$

We could find their sum, which is $\displaystyle 8221$ , and note that the units digit is $\displaystyle 1$ . However, we could find the units digit with far less calculation.

Solution

We can simply add the units digits of the summands:

$\displaystyle 3 + 1 + 8 + 9 = 21.$

The units digit of this sum is $\displaystyle 1$ , which must be the same as the units digit of the four-digit sum we computed earlier.

Why we only need to use remainders

We can rewrite each of the integers in terms of multiples of $\displaystyle 10$ and remainders:
$\displaystyle 2403 = 240 \cdot 10 + 3$
$\displaystyle 791 = 79 \cdot 10 + 1$
$\displaystyle 688 = 68 \cdot 10 + 8$
$\displaystyle 4339 = 433 \cdot 10 + 9$
When we add all four integers, we get

$(240 \cdot 10 + 3) + (79 \cdot 10 + 1) + (68 \cdot 10 + 8) + (433 \cdot 10 + 9)$ $= (240 + 79 + 68 + 433) \cdot 10 + (3 + 1 + 8 + 9)$

At this point we already see the units digits grouped apart and added to a multiple of $\displaystyle 10$ (which will not affect the units digit of the sum):

$= 820 \cdot 10 + 21 = 8200 + 21 = 8221$

Solution using modular arithmetic

Now let's look back at this solution, using modular arithmetic from the start. Note that
$\displaystyle 2403 \equiv 3 \pmod{10}$
$\displaystyle 791 \equiv 1 \pmod{10}$
$\displaystyle 688 \equiv 8 \pmod{10}$
$\displaystyle 4339 \equiv 9 \pmod{10}$
Because we only need the modulo $\displaystyle 10$ residue of the sum, we add just the residues of the summands:

$\displaystyle 2403 + 791 + 688 + 4339 \equiv 3 + 1 + 8 + 9 \equiv 21 \equiv 1 \pmod{10},$

so the units digit of the sum is just $\displaystyle 1$ .

Addition rule

In general, when $\displaystyle a, b, c$ , and $\displaystyle d$ are integers and $\displaystyle m$ is a positive integer such that

$\displaystyle a \equiv c \pmod{m}$ $\displaystyle b \equiv d \pmod{m}$

the following is always true:

$\displaystyle a + b \equiv c + d \pmod{m}$

.

And as we did in the problem above, we can apply more pairs of equivalent integers to both sides, just repeating this simple principle.

Subtraction

The same shortcut that works with addition of remainders works also with subtraction.

Problem

Find the remainder when the difference between $\displaystyle 60002$ and $\displaystyle 601$ is divided by $\displaystyle 6$ .

Solution

Note that $\displaystyle 60002 = 10000 \cdot 6 + 2$ and $\displaystyle 601 = 100 \cdot 6 + 1$ . So,
$\displaystyle 60002 \equiv 2 \pmod{6}$
$\displaystyle 601 \equiv 1 \pmod{6}$
Thus,

$\displaystyle 60002 - 601 \equiv 2 - 1 \equiv 1 \pmod{6},$

so 1 is the remainder when the difference is divided by $\displaystyle 6$ . (Perform the subtraction yourself, divide by $\displaystyle 6$ , and see!)

Subtraction rule

When $\displaystyle a, b, c$ , and $\displaystyle d$ are integers and $\displaystyle m$ is a positive integer such that

$\displaystyle a \equiv c \pmod{m}$ $\displaystyle b \equiv d \pmod{m}$

the following is always true:

$\displaystyle a - b \equiv c - d \pmod{m}$

.

Summary of Useful Facts

Consider four integers ${a},{b},{c},{d}$ and a positive integer ${m}$ such that $a\equiv b\pmod {m}$ and $c\equiv d\pmod {m}$ . In modular arithmetic, the following identities hold:

Addition: $a+c\equiv b+d\pmod {m}$ .
Subtraction: $a-c\equiv b-d\pmod {m}$ .
Multiplication: $ac\equiv bd\pmod {m}$ .
Division: $\frac{a}{e}\equiv \frac{b}{e}\pmod {\frac{m}{\gcd(m,e)}}$ , where $e$ is a positive integer that divides ${a}$ and $b$ .
Exponentiation: $a^e\equiv b^e\pmod {m}$ where $e$ is a positive integer.

The Natural Appeal of Modular Arithmetic

Observe that we use modular arithmetic even when solving some of the most basic, everyday problems. For example:

Cody is cramming for an exam that will be held at 2 PM. It is the morning of the day of the exam, and Cody did not get any sleep during the night. He knows that it will take him exactly one hour to get to school from the time he wakes up, and he insists upon getting at least five hours of sleep. At what time in the morning should Cody stop studying and go to sleep?

We know that the hours of the day are numbered from $1$ to $12$ , with hours having the same number if and only if they are a multiple of $12$ hours apart. So we can use subtraction mod $12$ to answer this question.

We know that since Cody needs five hours of sleep plus one hour to get to school, he must stop studying six hours before the exam. We can find out what time this is by performing the subtraction

$2 - 6 = -4 \equiv 8 \pmod{12}$ .

So Cody must quit studying at 8 AM.

Of course, we are able to perform calculations like this routinely without a formal understanding of modular arithmetic. One reason for this is that the way we keep time gives us a natural model for addition and subtraction in $\mathbb{Z}_n$ : a "number circle." Just as we model addition and subtraction by moving along a number line, we can model addition and subtraction mod $n$ by moving along the circumference of a circle. Even though most of us never learn about modular arithmetic in school, we master this computational model at a very early age.

Computation of Powers Mod n

The "exponentiation" property given above allows us to perform rapid calculations modulo $n$ . Consider, for example, the problem

What are the tens and units digits of $7^{1942}$ ?

We could (in theory) solve this problem by trying to compute $7^{1942}$ , but this would be extremely time-consuming. Moreover, it would give us much more information than we need. Since we want only the tens and units digits of the number in question, it suffices to find the remainder when the number is divided by $100$ . In other words, all of the information we need can be found using arithmetic mod $100$ .

We begin by writing down the first few powers of $7$ mod $100$ :

$7, 49, 43, 1, 7, 49, 43, 1, \ldots$

A pattern emerges! We see that $7^4 = 2401 \equiv 1$ (mod $100$ ). So for any positive integer $k$ , we have $7^{4k} = (7^4)^k \equiv 1^k \equiv 1$ (mod $100$ ). In particular, we can write

$7^{1940} = 7^{4 \cdot 485} \equiv 1$ (mod $100$ ).

By the "multiplication" property above, then, it follows that

$7^{1942} = 7^{1940} \cdot 7^2 \equiv 1 \cdot 7^2 \equiv 49$ (mod $100$ ).

Therefore, by the definition of congruence, $7^{1942}$ differs from $49$ by a multiple of $100$ . Since both integers are positive, this means that they share the same tens and units digits. Those digits are $4$ and $9$ , respectively.

A General Algorithm

In the example above, we were fortunate to find a power of $7$ -- namely, $7^4$ -- that is congruent to $1$ mod $100$ . What if we aren't this lucky? Suppose we want to solve the following problem:

What are the tens and units digits of $13^{404}$ ?

Again, we will solve this problem by computing $13^{404}$ modulo $100$ . The first few powers of $13$ mod $100$ are

$13, 69, 97, 61, 93, \ldots$

This time, no pattern jumps out at us. Is there a way we can find the $404^{th}$ power of $13$ mod $100$ without taking this list all the way out to the $404^{th}$ term -- or even without patiently waiting for the list to yield a pattern?

Suppose we condense the list we started above; and instead of writing down all powers of $13$ mod $100$ , we write only the powers $13^k$ , where $k$ is a power of $2$ . We have the following (all congruences are modulo $100$ ):

$\displaystyle 13^1 = 13$

$13^2 \equiv 69$

$13^4 \equiv 69^2 \equiv 61$

$13^8 \equiv 61^2 \equiv 21$

$13^{16} \equiv 21^2 \equiv 41$

$13^{32} \equiv 41^2 \equiv 81$

$13^{64} \equiv 81^2 \equiv 61$

$13^{128} \equiv 61^2 \equiv 21$

$13^{256} \equiv 21^2 \equiv 41$

(Observe that this process yields a pattern of its own, if we carry it out far enough!)

Now, observe that, like any positive integer, $404$ can be written as a sum of powers of two:

$404 = 256 + 128 + 16 + 4$

We can now use this powers-of-two expansion to compute $\overline{13}^{404}$ :

$13^{404} = 13^{256} \cdot 13^{128} \cdot 13^{16} \cdot 13^4 \equiv 41 \cdot 21 \cdot 41 \cdot 61 \equiv 61.$

So the tens and units digits of $13^{404}$ are $6$ and $1$ , respectively.

We can use this method to compute $M^e$ modulo $n$ , for any integers $M$ and $e$ , with $e > 0$ . The beauty of this algorithm is that the process takes, at most, approximately $2 \log_2 e$ steps -- at most $\log_2 e$ steps to compute the values $M^k$ modulo $n$ for $k$ a power of two less than $e$ , and at most $\log_2 e$ steps to multiply the appropriate powers of $M$ according to the binary representation of $e$ .

This method can be further refined using Euler's Totient Theorem.

Applications of Modular Arithmetic

Modular arithmetic is an extremely flexible problem solving tool. The following topics are just a few applications and extensions of its use:

Resources

The AoPS Introduction to Number Theory by Mathew Crawford.
The AoPS Introduction to Number Theory Course. Hundreds of students have learned more about modular arithmetic and problem solving from this 12 week class.

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Modular arithmetic/Introduction

Contents

The Basic Idea

Congruence

Examples

Sample Problem

Solution:

Making Computation Easier

Addition

Problem

Solution

Why we only need to use remainders

Solution using modular arithmetic

Addition rule

Subtraction

Problem

Solution

Subtraction rule

Summary of Useful Facts

The Natural Appeal of Modular Arithmetic

Computation of Powers Mod n

A General Algorithm

Applications of Modular Arithmetic

Resources

See also