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+k1)C(k1)
Positive solutions = (n1)C(k1)
Example : a+b+c+d = 10.
total non negative solution
= (10+41)C(41)
= 13C3
= 13x12x11/1x2x3
= 286.
positive solution
= (101)C(41)
= 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) (11/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 = 971 = 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+31)C(31)
= 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
= (136211)/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 = 2T1.
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.
= > (x3x8 )(y3x7) = 7x8x3^2.
= > (x24)(y21) = 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 = 2T1.
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.
(x6) (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 = 5.
negative solution = 6.
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.
(3x) (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 = 5
negative solution = 6.