# Concept: Factors

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CONTENTS

**INTRODUCTION**

A Factor of a given number completely divides it. **Example**: Factors of 6 are 1, 2, 3 and 6.

i.e., 6 is completely divisible by 1 or 2 or 3 or 6.

- A given Natural number will always have finite factors.

**PRIME FACTORISATION**

Prime Factorisation is a way of writing the given number as a product of powers of prime numbers.

Prime Factorised for of 12 = 2^{2} × 3

Prime Factorised for of 72 = 2^{3} × 3^{2}

**Example**: Prime Factorise 180.

**Solution:**

Given, 180 = 4 × 45

⇒ 180 = 2^{2} × 9 × 5

⇒ 180 = 2^{2} × 3^{2} × 5

∴ Prime Factorised form of 180 = 2^{2} × 3^{2} × 5

**NUMBER OF FACTORS of N**

If we need to find the number of factors of 72, we first prime factorise it as 2^{3} × 3^{2}

Now, any factor (f) of 72 will contain powers of either 2 or 3.

The power of 2 in f can be either 0, 1, 2 or 3, i.e., one more than the higest power of 2 in the number

The power of 3 in f can be either 0, 1 or 2, i.e., one more than the higest power of 3 in the number

∴ Number of factors that can be formed = (3 + 1)(2 + 1) = 12

**Generalising**

If prime factorised form of a number N = a^{p} × b^{q} × c^{r}

where a, b and c are prime numbers.

Then, Total number of factors of N = (p + 1)(q + 1)(r + 1)

**Example**: Find the number of factors of 180.

**Solution:**

180 can be factorised as 2^{2} × 3^{2} × 5^{1}

∴ Number of factors of 180 = (2 + 1)(2 + 1)(1 + 1) = 18

- Total number of factors of a perfect square will always be odd.
- Total number of factors of a non-perfect square will always be even.

Since p, q and r are even, (p + 1)(q + 1)(r + 1) will be odd.

Since at least one of p, q and r is odd, (p + 1)(q + 1)(r + 1) will be even.

**WRITING N AS A PRODUCT OF TWO OF ITS FACTORS**

Let us taken an example of 12. 12 can be factorised as 2^{2} × 3.

Factors of 12 are 1, 2, 3, 4, 6 and 12.

Now, we need to find number of ways of writing 12 as a product of two of its factors.

Product of 1 and 12 will give us 12, product of 2 and 6 will given us 12 and product of 3 and 4 will give us 12.

∴ Number of ways of writing 12 as a product of two of its factors is 3.

In general product of two factors gives us the required number, hence number of ways of writing a given number as a product of two of its factors is half the number of its factors.

**Generalising**

If N = a^{p} × b^{q} × c^{r}

Number of ways of writing N as a product of two of its factors = $\frac{1}{2}$(p + 1)(q + 1)(r + 1)

**Example**: In how many ways can we write 180 as a product of two of its factors.

**Solution:**

180 can be factorised as 2^{2} × 3^{2} × 5^{1}

∴ Number of ways of writing 180 as a product of two of its factors = $\frac{1}{2}$(2 + 1)(2 + 1)(1 + 1) = 9

**WRITING A PERFECT SQUARE AS A PRODUCT OF TWO OF ITS FACTORS**

Number of factors of a perfect square is always odd, hence dividing it by 2 will not give us the number of ways of writing it as a product of two of its factors. Let us understand perfect square with an example.

Let us take the example of 36. Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36 i.e., total 9 factors

Now,

36 = 1 × 36

36 = 2 × 18

36 = 3 × 12

36 = 4 × 9

Since 36 is a perfect square, one its factors is 6 (square root of 36).

36 = 6 × 6

∴ Number of ways of 36 as a product of two distinct factors = $\frac{1}{2}$(9 - 1) = 4

∴ Number of ways of 36 as a product of any two factors = $\frac{1}{2}$(9 - 1) + 1 (square of the square root) = 5

**Generalising**

If N (perfect square) = a^{p} × b^{q} × c^{r}

Number of ways of writing N as a product of two distinct factors = $\frac{1}{2}$[(p + 1)(q + 1)(r + 1) - 1]

Number of ways of writing N as a product of any two factors = $\frac{1}{2}$[(p + 1)(q + 1)(r + 1) - 1] + 1

**Example**: In how many ways can we write 144 as a product of two of its factors.

**Solution:**

144 can be factorised as 2^{4} × 3^{2}

∴ Number of ways of writing 144 as a product of two of its factors = $\frac{1}{2}$[(4 + 1)(2 + 1) - 1] + 1 = 8

**SUM OF FACTORS OF N**

Let us take an example of 12. 12 can be factorised as 2^{2} × 3.

Factors of 12 are

1 = 2^{0} × 3^{0}

3 = 2^{0} × 3^{1}

2 = 2^{1} × 3^{0}

6 = 2^{1} × 3^{1}

4 = 2^{2} × 3^{0}

12 = 2^{2} × 3^{1}

Sum of all factors of 12 = 2^{0} × 3^{0} + 2^{0} × 3^{1} + 2^{1} × 3^{0} + 2^{1} × 3^{1} + 2^{2} × 3^{0} + 2^{2} × 3^{1}

= 2^{0}(3^{0} + 3^{1}) + 2^{1}(3^{0} + 3^{1}) + 2^{2}(3^{0} + 3^{1})

= (2^{0} + 2^{1} + 2^{2})(3^{0} + 3^{1})
= (1 + 2 + 4)(1 + 3) = 28

**Generalising**

If prime factorised form of a number N = a^{p} × b^{q} × c^{r}

Sum of all factors of N = (a^{0} + a^{1} + ... + a^{p}) × (b^{0} + b^{1} + ... + b^{q}) × (c^{0} + c^{1} + ... + c^{r})

Sum of all factors of N = $\left(\frac{{\mathrm{a}}^{\mathrm{p}+1}-1}{\mathrm{a}-1}\right)$ × $\left(\frac{{\mathrm{b}}^{\mathrm{q}+1}-1}{\mathrm{b}-1}\right)$ × $\left(\frac{{\mathrm{c}}^{\mathrm{r}+1}-1}{\mathrm{c}-1}\right)$

**NUMBER OF WAYS OF WRITING N AS A PRODUCT OF CO-PRIMES**

**Generalising**

If prime factorised form of a number N = a^{p} × b^{q} × c^{r}

Number of ways of writing N as a product of 2 co-prime numbers = 2^{m-1}

where m is the number of distinct prime factors of N.

**Example**: 120 can be written as 3 × 40, 8 × 15, 24 × 5 and 120 × 1

We can write 120 as 2^{3} × 3 × 5

Thus, 120 has 3 distinct prime factors i.e., 2, 3 and 5.

∴ Number of ways of writing 120 as a product of 2 co-prime numbers = 2^{3-1} = 4.

**NUMBER OF CO-PRIMES TO N LESS THAN N**

**Generalising**

If prime factorised form of a number N = a^{p} × b^{q} × c^{r}

Number of co-primes to N which are less than N = N$\left(1-\frac{1}{\mathrm{a}}\right)$$\left(1-\frac{1}{\mathrm{b}}\right)$$\left(1-\frac{1}{\mathrm{c}}\right)$

**Example**: Let us taken the example of 12. Numbers which are less than 12 and co-prime to 12 are 1, 5, 7 and 11 i.e., 4 co-primes

We can calculate this using the above relation.

i.e., Number of co-primes to 12 less than 12 = 12$\left(1-\frac{1}{2}\right)$$\left(1-\frac{1}{3}\right)$ = 4

**SUM OF CO-PRIMES TO N LESS THAN N**

**Example**: Let us taken the example of 12. Numbers which are less than 12 and co-prime to 12 are 1, 5, 7 and 11 i.e., 4 co-primes

Co-primes to 12 less than 12 are 1, 5, 7 and 11.

Sum of 1 and 11 gives 12, while sum of 5 and 7 gives 12.

Hence, sum of 2 co-primes gives us the number itself. Hence sum of all co-primes to a number less than the number = number × half the number of co-primes.

**Generalising**

If prime factorised form of a number N = a^{p} × b^{q} × c^{r}

Sum of co-primes to N which are less than N = $\frac{{\mathrm{N}}^{2}}{2}$$\left(1-\frac{1}{\mathrm{a}}\right)$$\left(1-\frac{1}{\mathrm{b}}\right)$$\left(1-\frac{1}{\mathrm{c}}\right)$