Factors - Hemant Malhotra


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 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): 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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