Question Bank  Number of Integral/Positive/NonNegative Solutions  Set 1

Case 1
x = 2x  120 – 3x
x = 120 – 3x
case a) x = 120 – 3x
x = 30.
Case b) x = 3x – 120
x = 60.Case 2
x = 2x  120 – 3x
3x = 120 – 3x
3x = 120 – 3x
x = 20.x = 20, 30, 60.
3 values.

x^2 – y^2 = 627.
627 = 3 * 11 * 19.
Total number of factors = 2 * 2 * 2 = 8.
Total Integral solutions = 2 * 8 = 16

By direct formula,
x1 + x2 ... xK = N.
Positive solution = (N – 1)C(K1)
= (20 – 1)C(31)
= 19C2
= 171.

x1 + x2 + x3 + x4 = 20.
If a + b + c .... r = k
Then by total positive solution = (k1)C(r1)
Total positive solution = (201)C(41) = 19C3 = 969.But we need to remove case when x1 = x2.
x1 + x1 + x3 + x4 = 20.
2x1 + x3 + x4 = 20.
Total solutions = 17+15+13+11+9+7+5+3+1 = 81.This will be removed from total solutions = 969 – 81 = 888.
now only possibility in above equation is that x1 < x2 and x1 > x2.
So we need to take only x1 > x2 cases.
Hence total positive solution = 888 / 2 = 444.

x = 0, 1, 2, 1, 2, 3.
Total solution = 6 values.

Unit digit of N^4 = 1, 6, 5, where N is any natural number.
Unit digit of Sum of w^4, x^4, y^4, z^4 can never be 7.
Hence 0 solution.

Case 1
x^2 2x8 ≥ 0.
x^2 – 2x 8 = 0.
x = 4,2.
x^2 – 2x – 8 = 9.
x = 1.Case 2
x^2 + 2x – 8 ≥ 0.
x^2 + 2x – 8 = 0.
x = 4,2.
x^2 + 2x – 8 = 9.
x^2 +2x +1 = 0.
x = 1.Case 3
x^2 – 2x 8 = 8.
x= 0,2.For values of x = 4, 2, 1, 0, 1, 2, 4.
Total values = 7.

287 = 7 * 41
Total no of factors = 2 * 2 = 4
Total integral solution = 2 * 4 = 8

x^2 + y^2 = 100
Here, 100 is perfect square.
Number of ordered positive integral solutions = Number of factors of N minus 1 ignoring the presence of (4k + 3) primes and 2
Total integral solutions = (Ordered Positive integral solutions) * 4 + 4100 = 2^2 * 5^2
Number of ordered positive integral solutions = 3  1 = 2
Total Integral = 2 * 4 + 4 = 12solutions are  (0,10)(10,0)(10,0)(0,10),(6,8)(6,8),(6,8),(6,8), (8,6),(8,6), (8,6),(8,6).

From Q15 and Q14 we know the cases for x^2 + y^2 = 36 and ≤ 36.
Cases (< 36) = Cases (≤ 36)  Cases (= 36) = 113  4 = 109 ways.

(a+2) + (b+0) + (c5) + (d+8) = 30.
a + b + c + d = 25.
Total whole number solutions = (25 + 4  1)C(4  1) = 28C3 = 3276.

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]