# Question Bank - Number of Integral/Positive/Non-Negative Solutions - Set 1

• abc = 120.
120 = 2^3 * 3 * 5.
Here, positive solution = 5C2 * 3C2 * 3C2
= 10 * 3 * 3
= 90

• x = -8 to 12.
Total 21 values.

• Same as Q13.
36 = 2^2 * 3^2
Number of positive integral solutions = 0
Total integral solutions = 0 * 4 + 4 = 4
Solutions are (0,6), (0,-6), (-6,0), (6,0)

• It's best to use gauss's formula
Number of integral solutions of x^2 + y^2 ≤ r^2 = 1 + 4[r] + 4 * [(Summation i = 1 to R] √(r^2 - i^2)]
for x^2 + y^2 ≤ 6^2, integral solutions = 1 + 24 + 4 (5 + 5 + 5 + 4 + 3 + 0)
= 1 + 24 + 4 * 22
= 113

Note - This would be approximately equal to the area of the circle
x^2 + y^2 ≤ 6^2 (Circle of radius 6)
Integral solutions = [Area] = [3.14 x 36] = 113 (Approximation)

• Total - (0,0), (0,1),(1,0), (2,1), (1,2), (2,2)
Removing cases with 0, positive integral solutions = 3

• a^2 = 9555^2 + b^2
a^2 - b^2 = 9555^2
RHS is odd so positive solutions = (Number of factors of RHS)/2

9555^2 = 3^2 * 5^2 * 7^4 * 13^2
Number of factors = 3 * 3 * 5 * 3 = 135

positive solutions = [135/2] = 67

• Concept : a/x - b/y = 1/k.
T = Factors of (a * b * k^2)
Total Integer solution = 2T - 1.
positive solution = (T - 1)/2
negative solution = 0

Here, a = b = 1 and k = 12
T = Factors of 12^2 (= 3^2 * 2^4)
= 3 * 5 = 15
positive solutions = (15 - 1)/2 = 7

• 7x + 3y = 2000
put x = 2, y = (2000 - 14)/3 = 662
increase x as the co-efficient of y (3) [y will decreasing as the co-efficient of x]
x = 5, y = 655
x = 8, y = 648
x = 11, y = 641 and so on.
go on like this and we can reach till x = 284, y = 4
Total 95 solutions.

• 2x + 2y + z = 1000
x + 2y + 3z = 1000
x- 2z = 0
x = 2z
So 2y + 5z = 1000
+ve soln so z = 1 to 199
But 5z must b even else y will not be an integer
So 199-1/2=99 solutions
[Credits : @Learn-Quest]

• x^2 + y^2 - xy = x + y
(x - y)^2 + (x -1)^2 + (y -1)^2 = 2
(0,0) (0,1) (1,0) (2,2) (1,2) (2,1)
6 integral solutions
3 positive integer solutions
[credits : @Raman_Sharma]

45

61

61

1

1

1

92