Question Bank  Number of Integral/Positive/NonNegative 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
= 113Note  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)/29555^2 = 3^2 * 5^2 * 7^4 * 13^2
Number of factors = 3 * 3 * 5 * 3 = 135positive 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 = 0Here, 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 coefficient of y (3) [y will decreasing as the coefficient 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 1991/2=99 solutions
[Credits : @LearnQuest]

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]