# Difference between revisions of "Modular arithmetic/Introduction"

Aequilipse (talk | contribs) (→Problem #3) |
AlcumusGuy (talk | contribs) (→Congruence) |
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<center><math>311 \equiv 3 \pmod{4}</math></center> | <center><math>311 \equiv 3 \pmod{4}</math></center> | ||

and <math>3</math> is the modulo <math>4</math> residue of <math>311</math>. | and <math>3</math> is the modulo <math>4</math> residue of <math>311</math>. | ||

+ | |||

+ | [hide="Solution"] | ||

+ | Clearly, if <math>n = 0</math>, the sum is <math>3</math>, which is divisible by <math>3</math>, so <math>0</math> works. | ||

+ | |||

+ | Now let <math>n \in \matbb{N}</math> to get the following | ||

+ | |||

+ | <math>2^n + 3^n + 4^n \equiv 0 \pmod{3} \implies 2^n + 0 + 4^n \equiv 0 \pmod{3} \implies 2^n + 4^n \equiv 0 \pmod{3}</math> | ||

== Making Computation Easier == | == Making Computation Easier == |

## Revision as of 08:34, 17 March 2012

**Modular arithmetic** is a special type of arithmetic that involves only integers. This goal of this article is to explain the basics of modular arithmetic while presenting a progression of more difficult and more interesting problems that are easily solved using modular arithmetic.

## Contents

## Motivation

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

This is the way in which we count in **modulo 12**. When we add to , we arrive back at . The same is true in any other modulus (modular arithmetic system). In modulo , we count

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

where is the same as in modulo 5. Because all integers can be expressed as , , , , or in modulo 5, we give these integers their own name: the **residue classes** modulo 5. In general, for a natural number that is greater than 1, the modulo residues are the integers that are whole numbers less than :

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 : In other words, this means in base 5, these integers have the same last digit:

2(base 5) 12(base 5) 22(base 5) 32(base 5) 42(base 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:

In general, two integers and are congruent modulo when is a multiple of . In other words, when is an integer. Otherwise, , which means that and are **not congruent** modulo .

### Examples

- because is a multiple of .

- because , which is an integer.

- because , which is not a multiple of .

- because , which is not an integer.

### Sample Problem

Find the modulo residue of .

#### Solution:

Since R , we know that

and is the modulo residue of .

[hide="Solution"] Clearly, if , the sum is , which is divisible by , so works.

Now let $n \in \matbb{N}$ (Error compiling LaTeX. ! Undefined control sequence.) to get the following

## 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 , then we can work directly with those remainders in modulo . This can be more easily understood with a few examples.

### Addition

#### Problem

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

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

#### Solution

We can simply add the units digits of the addends:

The units digit of this sum is , 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 and remainders:

.

When we add all four integers, we get

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

#### Solution using modular arithmetic

Now let's look back at this solution, using modular arithmetic from the start. Note that

Because we only need the modulo residue of the sum, we add just the residues of the summands:

so the units digit of the sum is just .

#### Addition rule

In general, when , and are integers and is a positive integer such that

the following is always true:

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

Proof of the addition rule:

Let , and for . Adding the two equations we get:

$\begin{eqnarray*} mk+ml&=&(a-c)+(b-d)\\ m(k+l)&=&(a+b)-(c+d) \end{enarray*}$ (Error compiling LaTeX. ! Missing \endgroup inserted.)

Which is equivalent to saying

### Subtraction

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

#### Problem

Find the remainder when the difference between and is divided by .

#### Solution

Note that and . So,

Thus,

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

#### Subtraction rule

When , and are integers and is a positive integer such that

the following is always true:

.

### Multiplication

Modular arithmetic provides an even larger advantage when multiplying than when adding or subtracting. Let's take a look at a problem that demonstrates the point.

#### Problem

Jerry has 44 boxes of soda in his truck. The cans of soda in each box are packed oddly so that there are 113 cans of soda in each box. Jerry plans to pack the sodas into cases of 12 cans to sell. After making as many complete cases as possible, how many sodas will Jerry have leftover?

#### Solution

First, we note that this word problem is asking us to find the remainder when the product is divided by .

Now, we can write each and in terms of multiples of and remainders:

This gives us a nice way to view their product:

We can already see that each part of the product is a multiple of , except the product of the remainders when each and are divided by 12. That part of the product is , which leaves a remainder of when divided by . So, Jerry has sodas leftover after making as many cases of as possible.

#### Solution using modular arithmetic

First, we note that

Thus,

meaning there are sodas leftover. Yeah, that was much easier.

#### Multiplication rule

When , and are integers and is a positive integer such that

The following is always true:

### Exponentiation

Since exponentiation is just repeated multiplication, it makes sense that modular arithmetic would make many problems involving exponents easier. In fact, the advantage in computation is even larger and we explore it a great deal more in the intermediate modular arithmetic article.

Note to everybody: Exponentiation is very useful as in the following problem:

#### Problem #1

What are the last two digits of if there are 1000 7s as exponents and only one 7 in the middle?

We could solve this problem using mods. This can also be stated as . After that, we see that 7 is congruent to -1 in mod 4, so we can use this fact to replace the 7s with -1s, because 7 has a pattern of repetitive period 4 for the units digit. is simply 1, so therefore , which really is the last digit.

#### Problem #2

What are the tens and units digits of ?

We could (in theory) solve this problem by trying to compute , 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 . In other words, all of the information we need can be found using arithmetic mod .

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

A pattern emerges! We see that (mod ). So for any positive integer , we have (mod ). In particular, we can write

(mod ).

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

(mod ).

Therefore, by the definition of congruence, differs from by a multiple of . Since both integers are positive, this means that they share the same tens and units digits. Those digits are and , respectively.

#### Problem #3

Can you find a number that is both a multiple of but not a multiple of and a perfect square?

#### Solution to Problem #3

No you cannot, rewritting the question we see that it asks us to find a number that satisfies this: .

Taking mod on both sides we find that , now all we are missing to show is that no matter what is, will never be a multiple of plus , so we work with cases:

This assures us that it is impossible to find such a number.

## Summary of Useful Facts

Consider four integers and a positive integer such that and . In modular arithmetic, the following identities hold:

- Addition: .
- Subtraction: .
- Multiplication: .
- Division: , where is a positive integer that divides and .
- Exponentiation: where is a positive integer.

## 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. Thousands of students have learned more about modular arithmetic and problem solving from this 12 week class.