Number Theory : Factor Theorem - Vikas Saini



  • If N is a composite number = a^p x b^q x c^r

    Here a,b and c are prime numbers and p,q and r are their power respectively.

    Number of factors = (p+1)(q+1)(r+1)

    Let find the number of factors of 60.

    60 = 2^2 x 3 x 5.

    No of factors = (2+1)(1+1)(1+1) = 3 x 2 x 2 = 12.

    Factors of  60 are 1,2,3,4,5,6,10,12,15,20,30,60.

    No. of factors of a prime no = 2

    Find the number of factors of 97.

    97 = 97^1

    No of factors = 1+1 = 2.

    If any number has total number of factors are 3 only.

    It means that number is a square of a prime number.

    Q.  If 11 < n < 600.

    For how many values of n, the no of factors are 3 ?

    Solution :-

    We know 24^2 = 576 and 25^2 = 625.

    3^2 = 9 and 4^2 = 16.

    3^2 < n  < 24^2.

    We need to work on prime numbers between 3 &  24.

    The numbers are 5,7,11,13,17,19,23.

    Total 7 numbers.

    Even factors & odd factors of any number

    If any composite number N = 2^a x p^b x q^c

    No of Even factors = a x (b+1) x (c+1)

    No of Odd factors = (b+1) (c+1)

    Q. Find the even & odd factors of 432.

    432 = 2^4 x 3^3.

    Total factors = (4+1) (3+1) = 20.

    No of Even factors = 4 x (3+1) = 16.

    No of Odd factors = 3+1 = 4.

    Factors of 432 = 1,2,3,4,6,8,9,12,16,18,24,27,36,48,54,72,108,144,216,432.

    How to find even factors

    (2,4,8,16) (1,3,9,27)

    Firstly multiply by 2

    2(1,3,9,27) = 2,6,18,54

    Then by 4

    4(1,3,9,27)=4,12,36,108.

    Then by 8

    8(1,3,9,27)=8,24,72,216.

    Then by 16

    16(1,3,9,27)=16,48,144,432.

    Even factors are 2,6,18,54,4,12,36,108,8,24,72,216,16,48,144,432.

    How to find odd factors

    1,3,9,27 itself.

    Q.  Find total number of factors,even factors and odd factors of 300.

    Solution :-

    300 = 2^2 x 3 x 5^2.

    Total factors = (2+1) (1+1) (2+1) = 3 x  2 x 3 = 18.

    Even factors = 2 (1+1) (2+1) = 12.

    Odd factors = (1+1) (2+1) = 6.

    Multiple of 2 (1,2,4)

    Multiple of 3 (1,3)

    Multiple of  5 (1,5,25)

    To find Even factors :-

    (2,4)(1,3)(1,5,25)

    (2,4)(1,3)(1,5,25)

    (2,4)(1,5,25,3,15,75)

    (2,10,50,6,30,150,4,20,100,12,60,300)

    To find odd factors

    (1,3)(1,5,25)

    (1,5,25,3,15,75)

    Q. If 4n has 15  factors, 5n has 12 factors. Find no of factors in 3n.

    Solution :-

    Important part here is multiplication of 4 & 5.

    4 = 2^2.

    Let n = 2^a x 5^b.

    4n = 2^(a+2) x 5^b.

    Total factors = (a+2+1) (b+1)

    15 = (a+3) (b+1)…………………..(i)

    5n = 2^a x 5^(b+1)

    Total factors = (a+1) (b+1+1)

    12 = (a+1) (b+2)…………………….(ii)

    From equation (i) & (ii)

    a = 2 , b =2.

    n = 2^2 x 5^2

    3n = 2^2 x 3 x 5^2

    Total factors = 3 x 2 x 3 = 18.

    Q. For how many positive values of N,  it must be an integer

    (N + 102) / (N +2).

    Solution :-

    For such questions we need to work on factors only.

    (N+2+100) / (N+2)

    1 + 100 / N+2.

    100 = 2^2 x 5^2.

    Total factors = (2+1) (2+1) = 9.

    But here N+2 > 2.( N must be positive)

    We need to remove 1 & 2 from total factors.

    Required values = 9 – 2 = 7.

    Check :-

    Factors of 100 are 1,5,25,2,10,50,4,20,100.

    After removing 1,2 it can be checked by putting N+2 = 5,25,10,50,4,20,100.

    N=3,23,8,48,2,18,98.

    (N+102) / (N+2) = 21,5,11,3,26,6,2. (Total 7 values)

    Q. For how many total values of N, this must be an integer.

    (N + 28 ) / (N + 8 ).

    Solution :-

    We need to work on factors.

    (N+8+20) / (N+8 )

    1 + (20 / N+8 )

    60 = 2^2 x 5.

    Total factors = 3 x 2 = 6.

    But in question it is given that for total values, so factors can be negative too.

    Total = 2 x 6 = 12 .

    Check :-

    Factors of 20: – (1,2,4,5,10,20).

    Negative factors :- (-1,-2,-4,-5,-10,-20).

    N + 8 = -1,-2,-4,-5,-10,-20,1,2,4,5,10,20.

    N = -9,-10,-12,-13,-18,-28,-7,-6,-4,-3,2,12.

    (N+28 ) / (N+8 ) = -19,-9,-4,-3,-1,0,21,11,6,5,3,2.

    Total 12 integer values.

    Q.  Find the smallest number with 12 divisors.

    Solution :-

    12 = 2 x 6, 3x 4.

    To get smallest number we need to work on 2 & 3 only.

    N = 2^a x 3^b.

    Total divisors = (a+1)(b+1)

    (a,b) = (1,5)(5,1)(2,3)(3,2)

    N = 2 x 3^5 = 243.

    N = 2^5 x 3 = 96.

    N = 2^2 x 3^3 = 108.

    N = 2^3 x 3^2 = 72.

    Smallest number here is 72.

    Q. A composite number is given 150.

    (a) Find the sum of divisors.

    (b) Find the sum of even divisors.

    (c) Find the sum of odd divisors.

    Solution :-

    150 = 2 x 3 x 5^2.

    (a)Sum of divisors

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

    = (1+2)(1+3)(1+5+25)

    =3 x 4 x 31

     = 372.

    Check :-

    1+150+2+75+3+50+5+30+6+25+10+15 = 372.

    (b) sum of even divisors

    2 (1+3)(1+5+25)

    =2 x 4 x 31

    =248.

    Check :-

    150+2+50+30+6+10 =248.

    (c)sum of odd divisors

    (1+3)(1+5+25)

    = 4 x 31

    = 124.

    Check :-

    1+75+3+5+25+15=124.

    Q. How many divisors of 1296 are perfect square ?

    Solution =

    1296 = 36^2 = (2^2 x 3^2)^2

    1296 = 2^4 x 3^4

    Perfect square = [ (4/2) + 1] [(4/2) + 1] = 3 x 3 = 9.

    To find perfect square :-

    (1,2,4,8,16,32)(1,3,9,27,81)

    (1,4,16)(1,9,81)

    (1,9,81,4,36,324,16,144,1296)

    Total 9.

    Q. How many divisors of 36^36 are perfect cubes ?

    Solution :-

    36 = 2^2 x 3^2.

    36^36 = 2^72 x 3^72.

    Perfect cubes = [ (72/3) + 1] [(72/3) + 1]

    = 25 x 25 = 625.

    Q . How many of the first 1200  natural numbers are not divisible by  any 2,3,5 ?

    Solution :-

    Solve it by Euler’s theorem

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

    E(N) = N (1 – 1/p)(1 – 1/q) (1 – 1/r)

    1200 = 2^4 x 3 x 5^2.

    Numbers not divisible by 2,3,5

    = 1200 (1 – 1/2)(1 – 1/3) (1 – 1/5)

    = 1200 (1/2)(2/3)(4/5)

    = 320.

    Q. How many divisors of 112 are divisible by 4,but not 8.

    Solution :-

    112 = 2^4 x 7.

    Factors = 5 x 2 = 10.

    (1,2,4,8,16)(1,7)

    (1,2,4,7,8,14,16,28,56,112)

    4 = 2^2,8=2^3.

    No divisible by 4 = (5-2) x 2 = 6.

    No divisible by 8 = (5-3) x 2 = 4.

    Divisors = (6 – 4) = 2.

    But in 8, 4 is already counted.

    Hence 1.

    Q. How many divisors of 243 are divisible by 3, but not 27.

    Solution :-

    243 = 3^5.

    (1,3,9,27,81,243)

    3=3^1, 27=3^3

    No divisible by 3 = (6-1) = 5.

    No divisible by 27 = (6-3)= 3.

    Divisors = (5 – 3)= 2.

     


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