# Factors and Multiples

## Concept of Factors and Multiples

Let A and B be any two natural numbers.

If A/B is a natural number, then it means that:

- A is a multiple of B
- B is a factor of A

In other words, we can say that factors of a number are all those numbers that completely divide the given number, i.e. those that leave no remainder.

Factor of a number cannot be greater than the number (infact the largest factor will be the number itself). Thus the factors of any number will lie between 1 and the number itself (both inclusive) and thus are limited.

E.g. Factors of 6 are: 1, 2, 3 and 6.

The smallest multiple will be the number itself and the number of multiples would be infinite. E.g. Multiples of 6 are: 6, 12, 18 … so on.

You may have noticed that every successive multiple appears as the sixth number after the previous.

##### Q. Can you find the number of multiples of 6 less than 62?

Explanation:

We just need to find out the quotient when we divide 150 by 6 (and ignore the remainder).
So, 62/6 = 10

Hence, there are 10 multiples of 6 less than 62.

We can check it by writing all these multiples: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60.

## Factorization: Prime Factors

- Factors of a Number: As we have already seen, Factors of a number refers to all the numbers which divide that number completely (i.e. they leave no remainder).

E.g. These are the factors of 12: 1, 2, 3, 4, 6, 12.

The number is the multiple of its factors/divisors.

- Prime factors: The factors that are prime numbers.

E.g. These are the prime factors of 12: 2, 3.

### What is Prime factorization?

Prime factorization of a number is the expression of the number as the product of its prime factors.

N = $p^a q^b r^c$

Where, p, q and r are prime factors of the number, a, b and c are non-negative powers.

For example, This is the factorized form of 12 in terms of its basic prime factors: 12 = $2^2 × 3^1$

### Finding Number of factors

If N = $p^a q^b r^c$

(Where, p, q and r are prime factors of the number, a, b and c are non-negative powers.)

Then, Number of factors of N = (a + 1) (b + 1) (c + 1)

(including 1 and N)

E.g. 60 = $2^2 × 3^1 × 5^1$

So, possible powers of 2: (0, 1, 2)

Possible powers of 3: (0, 1)

Possible powers of 5: (0, 1)

Each unique combination of these powers gives a distinct factor.

The number of possible combinations = 3 × 2 × 2 = 12

Since there are 12 different combinations of the powers of 2, 3 and 5, there are 12 distinct factors of 60.

We will get the same answer if we apply the formula given above.

##### Q. Find the number of factors of the product $2^5 × 3^6 × 5^2$, which are perfect squares.

Explanation:

Any factor of this expression should be of the form $2^a × 3^b × 5^c$.

For the factor to be a perfect square a, b, and c have to be even.

a can take these values: 0, 2, 4

b can take these values: 0, 2, 4, 6

c can take these values: 0, 2

So, Total number of perfect square factors of the given product = 3 × 4 × 2 = 24

### Finding Number of odd factors

To find the number of odd factors we should ignore the power of the even prime factor (i.e. 2).

E.g. 60 = $2^2 × 3^1 × 5^1$

We will ignore the power of 2.

So, the number of odd factors of 60 = (1 + 1) (1 + 1) = 4

These are: 1, 3, 5, 15

### Finding number of even factors

Number of even factors = Total number of factors – Number of odd factors.

E.g. 60 = $2^2 × 3^1 × 5^1$

So, the number of even factors of 60 = 12 – 4 = 8

These are: 2, 4, 6, 10, 12, 20, 30, 60

## Sum and Product of all factors

### Product of all factors

Product of factors of N = $N^\frac{Number \hspace{1ex} of \hspace{1ex} factors}{2}$

(including 1 and the number itself)

For example:

12 = $2^2 × 3^1$

Number of factors of 12 = (2 + 1) (1 + 1) = 3 × 2 = 6

Product of the factors of 12 = $12^\frac{6}{2}$ = $12^3$

Let us double check this:

Product of the factors of 12 = 1 × 2 × 3 × 4 × 6 × 12 = $12^3$

### Sum of all factors

Sum of factors = $(p^0 + p^1 + … + p^a) (q^0 + q^1 + … + q^b) (r^0 + r^1 + … + r^c)$ …

= $\frac{(p^{a + 1} − 1)}{(p − 1)} \frac{(q^{b + 1} − 1)}{(q − 1)} \frac{(r^{c + 1} − 1)}{(r − 1)}$ …

(including 1 and the number itself)

For example:

12 = $2^2 × 3^1$

Sum of the factors of 12 = $\frac{(2^{2 + 1} − 1)}{(2 − 1)} \frac{(3^{1 + 1} − 1)}{(3 − 1)}$ = (7 × 8) / (1 × 2) = 28

Let us double check this:

Product of the factors of 12 = 1 + 2 + 3 + 4 + 6 + 12 = 28

#### Perfect numbers

If the sum of all factors (except that number itself) of a number is equal to the number itself is called a perfect number

E.g. Factors of 6 except 6 are: 1, 2, 3

Sum of above factors = 1 + 2 + 3 = 6

28 and 496 are the other perfect numbers in the first 1000 natural numbers.

## Number of ways to express a number as the product of two factors

There are two possible cases:

The number is a non-perfect square

The number is a perfect square

### Number is a non-perfect square

For such a number, the number of factors should be even.

The number of ways to express such a number as the product of two factors = $\frac{number \hspace{1ex} of \hspace{1ex} factors}{2}$

For example:

24 = $2^3 × 3^1$

The number of factors of 24 = (3 + 1) (1 + 1) = 8

So, the number of ways to express it as the product of two factors = $\frac{number \hspace{1ex} of \hspace{1ex} factors}{2}$ = 8/2 = 4

Let us double check it.

The number 24 can be expressed as the product of two factors in the following ways: 1 × 24, 2 × 12, 3 × 8, 4 × 6. (i.e. 4 ways)

### Number is a perfect square

For such a number, the number of factors should be odd.

Then the number of ways to express such a number as the product of any two factors = $\frac{number \hspace{1ex} of \hspace{1ex} factors + 1}{2}$ (including √Number x √Number)

And the number of ways to express the number as the product of two distinct factors = $\frac{number \hspace{1ex} of \hspace{1ex} factors - 1}{2}$ (excluding √Number x √Number)

For example:

25 = $5^2$

The number of factors of 25 = (2 + 1) = 3

So, the number of ways to express it as the product of two factors = $\frac{number \hspace{1ex} of \hspace{1ex} factors + 1}{2}$ = (3 + 1)/2 = 2

And the number of ways to express the number as the product of two distinct factors = $\frac{number \hspace{1ex} of \hspace{1ex} factors - 1}{2}$ = (3 - 1)/2 = 1

Let us double check it.

The number 25 can be expressed as the product of two factors in the following ways: 1 × 25, 5 × 5. (i.e. 2 ways, out of which one is the product of distinct factors)

## Number of ways to express a number as the product of two relatively prime factors

The number of ways to express a number as a product of two co-primes = $2^{n-1}$

(where n is the number of distinct prime factors of the number)

For example:

24 = $2^3 × 3^1$

Number of distinct prime factors of 24, n = 2 (i.e. 2 and 3)

So, the number of ways to express 24 as a product of two co-primes = $2^{n - 1}$ = $2^{2 - 1}$ = 2

These two ways are: 1 × 24, 3 × 8

## Properties of Factors

### Property 1: Factors of Prime Numbers

Perfect square of any prime number has exactly 3 factors.

E.g. Factors of 25: 1, 5 and 25.

### Property 2: Factors of Perfect Squares

All perfect squares have an odd number of factors.

E.g. Factors of 16: 1, 2, 4, 8 and 16, i.e. 5 factors

All non-perfect squares have even number of factors.

E.g. Factors of 15: 1, 3, 5, 15, i.e. 4 factors.

### Property 3: Divisibility by factors of a factor

If a number is divisible by another number, then it is also divisible by all factors of that number.

Example: 108 is divisible by 36

Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36

Hence, 108 is also divisible by each of the numbers 1, 2, 3, 4, 6, 9, 12, 18, 36.