Number of Integral, Non Negative & Positive Solutions - Vikas Saini



  • In this article, we will learn some basic and important variants of questions related to finding the number of solutions.

    Formulae reference - Indrajeet Singh (iQuanta)

    Integral solutions -- > solution can be any integer

    Positive integral solutions -- > solution can be any natural number

    Non Negative Integral Solutions -- > solution can be any whole number

    Type 1 : |x| + |y| = n

    Total solutions = 4n.

    Example : |x|+|y| = 5.

    Total solutions = 4 x 5 = 20.

    Type 2 : |x|+|y|+|z| = n.

    Total solutions = 4n^2 + 2.

    Example : |x|+|y|+|z| = 15.

    Total solutions = 4 x 15^2 + 2 = 902.

    Type 3 : |x|+|y|+|z|+|w| = n.

    Total solutions = (8/3)n(n^2 +2).

    Example : |x|+|y|+|z|+|w| = 9.

    Total solutions = (8/3) x 3 x (3^2 +2 )

    = 8 x 11

    = 88.

    ax+by = n.

    Non negative solutions = n/LCM (a,b) + 1. if either  a or b is divided by n.

    For positive solutions just remove x=0,y=0 from non negative solution.

    For n is not divisible by neither a nor b,then +1 either maybe at the end or maybe not.

    Example : 2x+3y = 30.

    positive integral solutions

    = 30 / LCM (2,3) - 1

    = 5 - 1 = 4.

    We have 4 solutions which are

    x = 3, y = 8.

    x = 6, y = 6.

    x = 9, y = 4.

    x = 12 , y = 2.

    Example : 2x+3y = 20

    Non negative integral solutions = 30 / LCM(2,3) + 1

    = 20/6 + 1 = 4.

    For x = 1, y = 6.

    x = 4, y = 4.

    x = 7, y = 2.

    x = 10, y = 0.

    Example : 3x+4y = 17

    positive solutions = 17/LCM(3,4)

    = 17/12 = 1.

    for x = 3, y = 2.

    a+b+c.............k terms = n.

    non negative integral solutions = (n+k-1)C(k-1)

    Positive solutions = (n-1)C(k-1)

    Example : a+b+c+d = 10.

    total non negative solution

    = (10+4-1)C(4-1)

    = 13C3

    = 13x12x11/1x2x3

    = 286.

    positive solution

    = (10-1)C(4-1)

    = 9C3

    = 9x8x7/3x2x1

    = 84.

    axb = N

    no of positive solution = no of factors of N.

    no of integrated solution =2 x  no of factors of N.

    Example : a x b = 36.

    36 = 2^2 x 3^2.

    total no of factors = (2+1)(2+1) = 9.

    positive solution = 9.

    total solution = 2 x 9 = 18.

    Here multiply by 2 because even negative sign also contains in total solution.

    Example : a x b = 27.

    27 = 3^3

    total no of factors = (3+1) = 4

    positive solution = 4

    (1,27),(3,9),(9,3),(27,1)

    total solution = 2 x 4 = 8.

    (-1,-27),(-3,-9),(-9,-3),(-27,-1),(1,27),(3,9),(9,3),(27,1).

    x + y = N.

    N = natural number.

    No of positive solution that x & y are co prime = E(N)

    if N = (a^x) x (b^y)

    E(N) = N (1 - 1/a) (1-1/b)

    Example : a+b = 46.

    no of positive solution a & b are co prime = E(46)

    46 = 2 x 23.

    E(46) = 46(1 - 1/2)(1- 1/23)

    = 46 (1/2) (22/23)

    = 22.

    a+b = P.

    P = prime number.

    no of positive solution for a & b are co prime = P - 1.

    Example : a+b = 97.

    no of positive solution = 97-1 = 96.

    N = x * y * z.

    N = (a^x) * (b^y)

    No of ordered positive solution = (x+2)C2 x (y+2)C2.

    No of integral ordered integral solution = 4 x (x+2)C2 x (y+2)C2.

    Q. In how many ways 72 can be written as

    (i) product of 3 positive integers

    (ii) product of 3 integers

    solution :-

    72 = 2^3 x 3^2.

    no of ordered solution = (3+2)C2 x (2+2)C2

    = 5C2 x 4C2

    = 10 x 6

    = 60.

    but we have to remove cases of aab & aaa cases.

    aab cases

    (i) 2^(0+0) b

    (ii) 2^(1+1) b

    (iii) (2x3)^(1+1) b

    (iv) 3^(1+1) b

    aab cases = 4.

    this 4 cases can be written in 3!/2! = 3 ways.

    aaa = 0 cases.

    total positive solution

    = ( 60 - 3 x 4) / 3! + 4

    = (60 - 12)/6 + 4

    = 48/6 + 4

    = 12.

    so 12 ways.

    (1,1,72),(2,2,18 ),(3,3,8 ),(6,6,2),(1,2,36),(1,3,24),(1,4,18 ),(1,6,12),(1,8,9),(2,3,12),(2,4,9),(3,4,6).

    (ii) total solution

    72 = 2^3 x 3^2

    total ordered solution = 4 x (3+2)C2 x (2+2)C2

    = 4 x 5C2 x 4C2

    = 4 x 10 x 6

    = 240.

    N = ABC

    N = (-A)(-B )C

    N = (-A)B(-C)

    N = A(-B )(-C)

    It means N can be writeen in 4 ways,this is why multiplied by 4.

    we need to remove aab & (-a)(-a)b.

    we have seen aab cases above.

    now (-a)(-a)b cases

    (i) (-1)^(0+0) b

    (ii) (-2)^(1+1) b.

    (iii) (-3)^(1+1) b.

    (iv) {(-2)(-3)}^2 b

    total cases = 4+4 = 8 cases.

    (240 - 3x8 )/3! + 8

    = (240 - 24)/6 + 8

    = 216/6 + 8

    = 36+8

    = 44.

    total = 44 ways.

    Q. In how many ways 3^15 can be written as product of 3 positive integers.

    solution :-

    3^15 = 3^(a+b+c)

    a+b+c = 15.

    total solution = (15+3-1)C(3-1)

    = 17C2

    = 17x16/2

    = 136.

    aab cases

    (0,0,15)(1,1,13),(2,2,11),(3,3,9),(4,4,7),(6,6,3),(7,7,1)

    aaa cases

    (5,5,5)

    we will remove both cases.

    so number of ways =

    (136 - 3x7 - 1)/3! + 7 + 1

    = (136-21-1)/6 + 8

    = 114/6 + 8

    =19+8

    = 27.

    Q. In how many ways 343000 can be written as product of 3 integers.

    solution :-

    343000 = 2^3 x 5^3 x 7^3

    total ordered solution =4 x (3+2)C2 x (3+2)C2 x (3+2)C2

    = 4 x (5C2)^3

    = 4 x (10)^3

    = 4 x 1000

    = 4000.

    aab cases :- 7

    (-a)(-a)b cases = 7

    (-a)(-a)a cases = 1

    aaa cases = 1.

    total solutions

    = (4000 - 3 x 7 - 3 x 7 - 3 x 1 - 1)/3! + 7 + 7 + 1 + 1

    = (4000 - 21 - 21 - 3 - 1)/3! + 16

    = (4000 - 46)/3! + 16

    = 3954/6 + 16

    = 659+16

    = 675.

    If question would have been asked just for positive integers.

    then ordered solution = (3+2)C2 x (3+2)C2 x (3+2)C2

    = 5C2 x 5C2 x 5C2

    = 1000.

    aab cases = 7

    aaa cases = 1

    total solution = (1000 - 3x7 - 1)/3! + 7+1

    = 978/6 + 8

    = 163+8

    = 171.

    a/x + b/y = 1/k.

    Total no of factors  = factors of (a x b x k^2) = T

    Total solution = 2T-1.

    Positive solution = T - 1.

    negative solution = 0.

    Q. 8/x + 7/y = 1/3.

    (i)Total solution

    (ii) Positive solution

    (iii) negative solution

    = > 3x8/x + 3x7/y = 1.

    = > (x-3x8 )(y-3x7) = 7x8x3^2.

    = > (x-24)(y-21) = 7x2^3x3^2.

    T = (1+1)(3+1)(2+1)

    T = 2 x 4 x 3 = 24.

    (i)Total solution

    = 2 x 24 - 1

    = 47

    here subtracted 1,because we have to remove x/y = 0 case.

    (ii) positive solution

    T - 1 = 24 - 1 = 23.

    (iii) negative solution

    = 0.

    a/x - b/y = 1/k.

    Total No of factors = axbxk^2 = T

    Total solution = 2T-1.

    positive solution = no of factors of T which are below a x k.

    Example : 2/x - 1/y = 1/3.

    total no of factors of (2 x 1 x 3^2)

    = (1+1)(2+1)

    = 6.

    Total solution = 2T - 1

    = 2 x 6 - 1

    = 11.

    Detailed explanation:-

    6/x - 3/y = 1.

    (x-6) (y+3) = -18.

    (6 - x) (3+y) = 18.

    here

    x = 5, y = 15.

    x = 4, y = 6.

    x = 3,y = 3.

    x = -3, y = -1

    x = -12, y = -2.

    x = 7, y = -21.

    x = 8, y = -12.

    x = 9, y = -9.

    x = 15, y = -5.

    x = 24, y = -4.

    x = 12, y = -6.

    So total solution = 11.

    positive solution = 3.

    negative solution = 8.

    Question : 1/x - 2/y = 1/3.

    total no of factors of 2x3^2 = 6.

    total solution = 2 x 6 - 1 = 11.

    Explanation :

    3/x - 6/y = 1.

    (3-x) (y+6) = 18.

    here

    x = 2, y = 12.

    x = 1, y = 3.

    x = -3, y = -3.

    x = -6, y = -4.

    x = -15, y = -5.

    x = 4, y = -24.

    x = 5, y = -15.

    x = 6, y = -12.

    x = 9, y = -9.

    x = 12, y =-8.

    x = 21, y= -7.

    positive solution = 2

    negative solution = 9.

     



  • WOW! Such an amazingg article :) But , In last two questions, according to formula positive solutions should be 3 and 2 respectively but they are given to be 5 and 5.Please see to this!



  • @CAT17 Thank you :)



  • Thanks for replying Sir.You completely cleared my confusion! please help me with this also: In last category of questions: no.1 (8/x)+(7/y)=1/3 total solutions are given to be 2T-1 =47 and positive solutions are T-1=23 so what are remaining soltions (47-23)?? if negative solutions are mentioned to be 0??


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