In this article, we will discuss Integers and their properties. We will discuss properties such as divisibility rules, multiples, factors, prime factors and the power of prime factorization, squares of integers, greatest common factor, least common multiple and the concepts of even and odd integers.

So let’s get started:

Contents

### Working with Integers

Integers are the non-fraction numbers including all the positive whole numbers, zero and negative whole numbers [ ..,-4,-3,-2,-1,0,1,2,3,4,…].

Integers have factors(also divisors) and multiples and these two will form the basis for our subsequent discussion.

If A is a **factor** of B, then B is a **multiple** of A. This also means that B is **divisible** by A.

### Divisibility and Divisibility Rules

Divisibility rules for integer can be used to determine whether the integer is a factor of another integer without doing extensive calculation or using the calculator. For example at a glance what will be the answer to the following questions:

Is 56 divisible by 7? yes.

Is 50 divisible by 13? no.

These are rather simple examples but what about a more difficult example:

Is 1234215675 divisible by 3?

We can use the divisibility rule to determine the answer without performing an extensive calculation.

Let us now consider some rules as they apply to certain integers:

##### Divisibility by 2

This is straight forward, all even numbers are divisible by 2 so if the last digit of a number(the one’s digit) is even, that number is automatically divisible by 2.

##### Divisibility by 3

To determine whether an integer is divisible by 3, we add up all the digits of the number. If the sum of the digits is divisible by 3, then the number is divisible by 3 otherwise the number is not divisible by 3.

E.g is 1296 by 3?

1+2+9+6 = 18 and since 18 is divisible by 3, then 1296 is divisible by 3.

Is 102,334,155 divisible by 3?

Summing all the digits 1+0+2+3+3+4+1+5+5 = 24, since 24 is divisible by 3, then 102,334,155 is divisible by 3.

##### Divisibility by 4

For whether a number is divisible by 4, we use the last 2 digits of the number(the ten’s place and the one’s place). If the last 2 digits form a 2-digit number divisible by 4, then the entire number is divisible by 4.

Consider the following 2 numbers and determine their divisibility by 4:

102,334,1* 55* -> 55 is last 2-digit number but not divisible by 4, therefore the whole number is not divisible by 4.

267,914,2* 96* -> 96 is divisible by 4, therefore the whole number is divisible by 4.

##### Divisibility by 5

Divisibility by 5 is straight forward and can be determined by looking at the last digits of a number. If the last digit is 0 or 5, then the number is divisible by 5.

E.g 102,324,145; 1,346,430; 43,567,890; 2,342,560 etc.

##### Divisibility by 6

This rule is a combination of the rules for 2 and 3. In order for a number to be divisible by 6, it has to be divisible by 2 and 3.

For example:

Is 1296 divisible by 6?

First, it’s even so it’s divisible by 2, also 1+2+9+6=18 and since 18 is divisible by 3, the number is also divisible by 3.

Since 1296 is both divisible by 2 and 3, it is therefore divisible by 6.

##### Divisibility by 9

This is exactly like the rule for 3. Add the digits together and if the sum is divisible by 9, then the number is divisible by 9 otherwise it is not divisible by 9. In the 1296 example, we have been using, the number is divisible by 9 since the sum of its digits is 18 which is divisible by 9.

How about 3072, sum = 3+0+7+2 = 12 which is not divisible by 9, therefore 3072 is not divisible by 9.

In summary, we have discussed the common divisibility rules for 2,3,4,5,6 and 9.

### Multiples

Under this subtopic, we will cover some basic concepts that relate to multiples:

##### Multiple Concept 1:

1 is a factor of every positive integer and so every positive integer is a multiple of 1. Just as every positive integer is a factor of itself, every positive integer is a multiple of itself.

To find the first n multiples of a number, we simply multiply the number by the numbers {1 … n} e.g first 5 multiples of 12 is {12,24,36,48,60}. We could also get the list by repeatedly adding 12.

##### Multiple Concept 2:

If we know that B is a multiple of A, then it must be true that B-A and B+A are also multiples of A.

For example, if 49 is a multiple of 7, then 49-7 = 42 and 49+7=56 are also multiples of 7.

##### Multiple Concept 3:

If Z and Y are multiples of X, then Z+Y and Z-Y are also multiples of X.

For example, since 700 and 49 are multiples of 7, therefore;

700+49=749, 749+49=798, 700-49=651 and 651-49=602 are all multiples of 7.

As a matter of fact, since 49 is a multiple of 7, 49+49+49+ … will always be a multiple of 7.

Therefore if B is a multiple of A, then any multiple of B is a multiple of A.

##### Multiple Concept 4:

If Z and Y are both multiples of X, then their product Z*Y must also be a multiple of X.

For example if 24 and 80 are both multiples of 8. then 24*80=1920 is also a multiple of 8.

Putting all the above concept together, let’s look at a problem;

If B, (B+200), (B+350) are all multiples of A, then A could be equal to which of the following: 20,25,75,100,150.

Using the principles we learn in concepts 2 and 3 above, we can derive additional multiples of A:

(B+200) – B = 200

(B+350) – B = 350

(B+350) – (B+200) = 150

And from these, we can derive one more multiple of A which is 200-150=50.

So 50, 150, 200 and 350 are all multiples of A, therefore, A must be a factor of all these numbers.

In the choice given, only 25 is a factor of the above numbers, therefore, 25 is the answer.

### Prime Numbers And Prime Factorization

A prime number is any positive integer that has only 2 factors, 1 and itself. The first couple of prime numbers are 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59.

Note that 1 is not a prime as it does not fit the definition above. 1 does not have two factors but one which is itself.

Also, note that 2 is the only even prime factor.

To test whether a number less than 100 is prime, all we have to do is check whether it is divisible by one of the prime numbers less than 10. If a number is not divisible by any prime number less than 10, then the number is itself a prime number.

##### Prime Factorization

Prime numbers are very important positive integers as they are the building blocks of all other positive integers. Why?

Every positive integer greater than 1 must either be a prime number or be expressed as a unique product of prime numbers.

In other words, every integer greater than 1 that is not prime can be expressed as a product of primes. That is to say, a number can be broken down as a product of only its prime factors.

This product is called the **prime factorization** of the number.

Let’s see some examples:

9 = 3*3

10 = 2*5

12 = 4*3 = 2*2*3

15 = 3*5

24 = 8*3 = 2*4*3 = 2*2*2*3

100 = 2*2*5*5

105 = 3*5*7

462 = 2*3*7*11

4680 = 2*2*2*3*3*5*13

The prime factorization of a number is like the DNA of the number through which we can see all the essential elements of the number.

For example, let us consider the number 96.

96 = 2*48 = 2*6*8 = 2*2*3*2*2*2

From the above, we can see what can and can’t be a factor of 96. Notice that 2,4,6,8,12,16,24,48 are all factors of 96 because their own prime factors are a subset of 96 prime factors.

Consider 16=2*2*2*2; there are five 2’s in 96, so 16 with four 2’s perfectly fit in as a factor.

But consider 18 = 2*3*3; there are two 3’s in 18 whereas there is only one 3 in 96 so 18 cannot fit in as a factor.

**Any factor of a number must itself be composed of only prime factors found in the number.**

Let us look at another example:

Given 4680 = 2*2*2*3*3*5*13

Which of the following numbers is or is not a factor of 4680?

25 = 5*5 NO

45 = 3*3*5 YES

65 = 5*13 YES

85 = 5*17 NO

120 = 2*2*2*3*5 YES

180 = 2*2*3*3*5 YES

This is just one way in which prime factorizations can be useful. Finding prime factorization is one of the most important things to know about Integers.

### Counting Factors of Large Numbers

Imagine given a large number say 345,890 and we are asked to determine how many factors it has. Rather than going through the long route of division, there is a simple trick we can use.

To find the number of factors that a positive integer N has:

- Find the prime factorization of N and write it in terms of powers of prime factors.
- Create a list of the exponents of the prime factors (use 1 for prime a factor that has no exponent i.e 2 has an exponent of 1 since 2 = 2 raised to the power of 1).
- Add one to every number on the list from step 2, creating a new list.
- Find the product of the new list. That product is the number of factors N has.

Let us consider an example:

How many factors does 21,600 have?

- 21600 = 216 * 100 = 2*108 *100 = 2*2*2*2*2*3*3*3*5*5 = 2
^{5}* 3^{3}* 5^{2} - List of exponents = {5,3,2}
- Adding 1 to each element = {6,4,3}
- Product of new list 6*4*3 = 72

There are 72 factors in 21,600.

To find the number of **odd factors**, we would basically repeat this procedure but ignore the exponent of the prime factor 2 when creating the list in the second step.

E.g for 21,600 = 2*2*2*2*2*3*3*3*5*5

Now we list the exponents of only odd prime factors, {3,2} ignoring the exponent of 2.

{3,2} -> {4,3} -> 4*3 = 12, there are 12 odd factors.

We have no direct way to calculate the number of even factors. But we can calculate the number of all the factors first and then calculate the number of all the odd factors and then subtract to get the number of even factors.

In our example, it will be 72-12 = 60 even factors.

### Squares of Integers

We can prime factorization to determine whether an integer is a perfect square of another integer.

Let’s use one perfect square as an example: 144

144 = 2*2*2*2*3*3 = 2^{4} * 3^{2} = (2*2*3) * (2*2*3) = 12*12

If the exponent of the prime factors of an integer are all even, then that integer is a perfect square.

A^{2} = A*A = (factors of A)*(factors of A)

This means that if we see an unknown number in its prime factorization form and all the exponents are even, we know the number must be a perfect square.

For example A = 2^{6} * 3^{4} * 5^{2}

This number A must be a perfect square since

A = 2^{6} * 3^{4} * 5^{2} = (2^{3} * 3^{2} * 5) * (2^{3} * 3^{2} * 5) = 360 * 360 = 360^{2}

##### Counting Factors in a Perfect Square

Since all the powers of prime factors of a perfect square are even, when we are counting factors, the list of exponents will be a list of all even numbers.

Then, when we add 1 to each number on the list, the second list will be a list of all odd numbers. This means the product will have to be odd.

Therefore, **a perfect square always has an odd number of factors.**

### Greatest Common Factor

The greatest common factor of **GCF** for short, sometimes all called the greatest common divisor, is simply the largest number on the common factor list of 2 or more numbers.

E.g to find GCF of 24 and 40, we find all their factors.

24: {1,2,3,4,6,8,12,24}

40:{1,2,4,5,8,10,20,40}

Common factors are {1,2,4,8}

The greatest GCF is 8.

We can always list the factors, identify the common ones and find the largest (GCF). The problem is this can become cumbersome and time-consuming when we are dealing with large numbers.

Instead, we can use a concept involving **prime factorization **to quickly find the GCF of numbers.

Let’s see an example: What is the GCF of 720 and 1200?

Prime Factorization:

720 = 2^{4} * 3^{2} * 5

1200 = 2^{4} * 3 * 5^{2}

All we have to do is find the highest power for each number’s prime factors common to both

The highest power of 2 common to both is 4 since both can fit in 2^{4}

The highest power of 3 common to both is 1 (they both have at most 1 3’s in common)

The highest power of 5 common to both is 1

GCF = 2^{4} * 3 * 5 = 240

### Least Common Multiple

The Least Common Multiple or **LCM **for short also known as Least Common Denominator is the lowest member of a set common the common multiples of two or more numbers.

E.g to find the LCM of 8 and 12.

Multiples of 8: {8,16,24,32,40,48,56,64,72 ….}

Multiples of 12: {12,24,36,48,60,72 …}

Common Multiples: {24,48,72 …}

And so the LCM is the lowest member of their common multiples = 24

The product of the 2 numbers is always one of the common multiples but not necessarily the LCM ofthe 2 numbers.

To find the LCM, we can always list the multiples and find the least common one but if the numbers are large, this process becomes time-consuming.

Prime factorization can make finding the LCM faster by finding the prime factorization and GCF first.

Let’s see an example: What is the LCM of 24 and 32?

Find the prime factorizations and the GCF.

24 = 2^{3} * 3

32 = 2^{5}

GCF = 8

Now write each number in the form GCF times another factor like so

24 = 8*3

32 = 8*4

The LCM is the product of these three factors.

LCM = 8*3*4 = 96.

In general, if A and B are 2 numbers and G is the GCF.

The LCM = G * A/G * B/G = A*B/G

In summary LCM = product of the two numbers divided by their GCF

LCM = (A*B)/GCF

What is the LCM of 12 and 75?

GCF = 3

LCM = 12*75/3 = 12*25=6*50 = 300

LCM can be useful when adding and substracting fractions since the LCM is also the Least Common Denominator.

Other important two things to note about LCM

- If A is a factor of B, then the LCM of A and B must be B. e.g LCM of 8 and 24 is 24.
- If it’s obvious that A and B have no factors in common greater than 1, then their LCM would simply be their product A*B.

### Even and Odd Integers

Even numbers go in both negative and positive directions:

……, -6,-4,-2,0,2,4,6,……

Also, Odd numbers go in both directions:

……, -5,-3,-1,1,3,5,…….

A few things to note here:

- Zero(0) is an even number.
- Evens and Odds include both positive and negative numbers.
- Evens and Odds pertain only to integers, any non-integer is neither even nor odd.
- All even numbers are divisible by 2. Even numbers can be expressed as 2n, where n is any integer. The prime factorization of a positive even integer greater than two has to absolutely include 2.
- No odd number is divisible by 2. Odd numbers can be expressed as (2n+1) where n is any integer. The prime factorization of a positive odd number will never contain a factor of 2.

Let us now consider some concepts of even and odd integers.

##### Adding and Subtracting Evens and Odds

E+E = E (Even + Even = Even) and E-E = E

E.g 2+2 = 4, 4+0 = 4,4-2 =2 etc

O+O = E (Odd + Odd = Even) and O-O = E

E.g 1+1 = 2, 3+5 = 8, 5-3 =2 etc

E+O = O and E-O = O

E.g 1+2=3, 2=3=5, 5-4=1 etc

In summary when we add or substract “likes(E and E or O and O)”, we get **Even**.

When we add and subtract “Unlikes (E and O)”, we get **Odd**.

##### Multiplying Evens and Odds

E*E = E

O*O = O

E*O = E

As long as there is at least one even factor in a product, the product will be even.

The only wat a product is odd is if every single factor is odd.

For example, if the product of five integers is even:

A*B*C*D*E = even

Any individual integer could be even or odd, all we know is that at least one of the five factors is even.

But if the product of the five integers is odd:

A*B*C*D*E = odd

Then we absolutely know that each one of the five factors is an odd number.

##### Dividing Evens and Odds

There are no general rules for odds and evens with division, simple because and integer divided by an integer might not even equal an integer in the first place.

When the quotient of two integers is another integer, whether that integer is even or odd depends on what factors cancelled and what factors remained.

It cannot be predicted simply by whether the two numbers divided were even or Odd.

E/E could be E or O or not even an integer at all.

O/O could be an O or not an integer at all.

E/O could be E or not an integer at all.

O/E is never an integer at all.

Let’s now consider an example with the above principles in action.

Given A, B, C and D are integers. If A is even and (A*B + C*D) is an odd integer, then which of the following must be true?

- B is odd
- C is odd
- D is odd

- Looking at (A*B + C*D), for this sum to be odd, one of the products must be odd and the other must be even (E+O = O or O+E = O).
- Since A is given as even and since (E*E = E or E*O = E), A*B must be even and therefore B can be either even or odd so
**B is odd is not always true.** - This also means the other product C*D is odd based on the first point.
- Now C*D is odd means both C and D are odds since only (O*O = O).

So only **C is odd must be true** and **D is odd must be true.**

### Conclusion

In this article, we have discussed some properties of integers and have seen the power of prime factorization and how it can help solve a lot of integer problems faster.

In the next article, we will discuss the powers and roots of numbers.