Factors  Hemant Malhotra

hemant_malhotra
Director at ElitesGrid  CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
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): 3^{4} * 5^{2} 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 = 2^{5 }* 3^{3 }* 5^{2 }
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=2^{4 }* 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)
= (2^41)/ (21) * (5^21)/ (51)
= 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 =324=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 = 2418=6