# Factors - Hemant Malhotra

• This Post consists of some basic Factors Questions :)

For a natural number N, all the numbers which can divide N completely are called divisors or factors of N

Let's count how many divisors 40 has. 1, 2, 4, 5, 8, 10, 20, 40 all divide 40 completely so number of divisors = 8 but if number is very large then it is really difficult to count them all. So there are some tricks and formulas for finding the number of factors of a given number.

First we will understand some basic stuffs: Let say you have 10 mango of same kind, 20 banana of same kind and 30 apple of same kind. Now you have options to select none or 1 or more than 1 items (any number of items)

CASE 1 -  (mango) we can select 0 mango or 1 mango or 2 mango or 3 mango and so on , so we have 11 ways to select mangoes (10+1)=11

CASE 2 - Banana, same (20+1) ways

CASE 3   - Apple same (30+1) = 31 ways

So total number of ways = (10+1) (20+1) (30+1)

Same idea we will apply in case of finding number of divisors

We break number in to their prime factors and then apply this theory

Example-

40 =2^3 * 5^1

Now we have 2 (3 times) and 5 (one time)

So for 2 we have 4 option, either to choose no 2 or one 2 or two 2, or three 2 so (3+1) =4 ways and for 5 we have 2 options

So (1+1) = 2 so total = (3+1) (1+1)

= 4 * 2 = 8

Let N be a composite number such that N= (p1)a * (p2)b * (p3)c , where p1,p2 ... pn are prime numbers. Then number of divisors of N = (a+1)*(b+1)*(c+1) …

Example: 70 = 2 * 5 * 7

Number of divisors= (1+1) (1+1) (1+1) = 8

QUESTION 1- Total number of factors of 2160

Approach

2160 = 2^4 * 3^3 * 5

Number of factors = (4+1) (3+1) (1+1) = 5 * 4 * 2 = 40

Question 2- Find number of even divisors of 5040

Approach

5040 =2^4 * 3^2 * 5 * 7 (even means number is a multiple of 2 so we must have at least one 2 so we cannot afford that case where we don't choose 2 so 5040 = 2^4 * 3^2 * 5 * 7.

So basically we want to select at least one 2 and any number of 3, 5 and 7 so number of ways (4)*(2+1)*(1+1)*(1+1) = 4*3*2*2 = 48

Question 3

N = 2^5 * 3^4 * 5^2

a) Find number of divisors

b) Find number of even divisors

c) Find number of odd divisors

d) Number of divisors whose unit digit is 5

e) Find number of divisors which is divisible by 10

f) Find number of divisors which is divisible by 6

Approach –

a) Number of divisor = (5+1) (4+1) (2+1)

b) Number  of even divisors means we need at least one 2 so (5)*(4+1)*(2+1) = 75

c) Number of odd means we don't need any two so (4+1) (2+1) = 15

d) Unit digit is 5 means we need at least one 5 and no 2 (because one 2 is there that will make unit digit 0): 34 * 52 so (4+1) * (2) = 10

e) Divisible by 10 means unit digit is 0 so should at least one 2 and at least one 5 so (5)*(4+1)*2=50

f) Divisible by 6 means at least one 2 and one 3 so (5)*4*(2+1)=60

Question 4 - how many divisors of 21600 are perfect square?

Approach –

21600 = 25 * 33 * 52

For perfect square, we need even powers of 2, 3 and 5

So 2^0, 2^2 and 2^4, 3^0 and 3^2, 5^0 and 5^2

So (3) * 2 * 2 = 12

Question 5 - Find number of divisors of form 4*n + 2 (n > =0) of 240

Approach

240=24 * 3 * 5

4*n+2 = 2*(2*n+1) =2*(odd number))

So we need only one 2 so number of divisors=1*(1+1) (1+1) = 4

Question 6 - How many divisors of 18^18 are perfect cube?

Approach

18^18=2^18*3^36 for perfect cube factors power should be multiple of 3

So 2^0 or 2^3 or 2^6 or 2^9 till 2^18

So 7 values

For 3, 3^0, 3^3 till 3^36 so 13 values

So 7*13=91 divisors which are perfect cube

SUM OF DIVISORS

Question 7 : Find sum of all divisors of 40

40=2^3*5 so we have options 2^0 or 2^1 or 2^2 or 2^3 and for 5, we have 5^0 or 5^1

So sum = (2^0+2^1+2^2+2^3) (5^0+5^1)

=15 * 6=90

Now understand this

40=2^3*5

So possible divisors:

2^0*5^0

2^0 *5^1

2^1*5^0

2^1*5^1

2^2*5^0

2^2*5^1

2^3*5^0

2^3*5^1

So these are the possibilities and we have to add them all to get the sum.

(2^0+2^1+2^2+2^3) (5^0+5^1)

= (2^4-1)/ (2-1) * (5^2-1)/ (5-1)

= 90

This is simple GP sum. so no need to remember direct formula

Question 8 - Find the no. of divisors of 1080 excluding the divisors which are perfect squares

Approach 1080=2^3*3^3*5 so total number of divisors=4*4*2=32

Now divisors which are perfect square will have even power of 2 and 3

So 2^0, 2^2 and 3^0, 3^2 so 2*2=4

So divisors which are not perfect square =32-4=28

Question 9 - How many factors of 360 are not divisors of 540?

Approach

360= 2^2*3^3*5 so number of factors=24

540=2^2*3^3*5 so number of factors= 24

Now common (HCF) =2^2*3^2*5

So number of common factors=3*3*2=18 and number of factors of 360 which are not factors of 540 = 24-18=6

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