# 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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