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

Q 20) How many pair of integers are possible for x^2  y^2 = 287?
OA : 8

Q 21) How many pair of positive integers are possible for a^2 = 9555^2 + b^2
OA: 67

Q 22) No of integral solution for (x3)^3 (x6)^4 (x9)^9 < 0 ?
OA : 4

Q 23) How many integral solutions does the equation mn – 2m = 3n + 105 have?
OA : 8

Q 24) Find the number of positive integral solutions of the equation 2x + 3y = 763?
OA : 127

Q 25) How many positive integral solutions are possible for 1/x  1/y = 1/12?
OA : 7

Q 26) How many positive integral solutions does equation a + b + c + d = a × b × c × d has
OA : 12

Q 27) Find the sum of integral roots of the following equation: 12x^4 − 91x^3 + 194x^2 − 91x + 12 = 0
OA : 7

Q 28 ) If m^3  n^3 = 9^k + 123. Find the number of nonnegative integral triplet(s) (m, n, k) satisfying the given equation.
OA : 1

Q 29) Find the total number of positive integral solution of 1/x + 1/y + 1/z = 1
OA : 10

Q 30) Number of nonnegative integral solutions of x^(2/3) + x^(1/3 ) – 42 ≤ 0
OA : 217

a + b + c ≤ 10 can be written as a + b + c + d = 10, where d is a dummy variable.
Why we need a dummy variable here ?
It is like we are dividing 10 items into 4 baskets and the fourth dummy basked (d) will ensure x, y and z would together have ≤ 10 always. If d = 0, a + b + c = 10 and if d = 10, a + b + c = 0 > covering all the possible values for a + b + c which are ≤ 10.
so the answer here is (10 + 4  1) C (4  1) = 13C3 = 286 ways

@zabeer
2x+3y=763
for y=1, x=380
from here x will change by coefficient of y i.e.(3) while y will change by coefficient of x i.e (2) in opposite or one will increase while another will decrease.
now, lets get back to there
y=1, x=380
y=3, x=377
y=5, x=374
and so on till
y=253 , x=2
total value= (3802)/3 +1=127

@zabeer
For x^2  y^2 = n
Where n is odd and not a perfect square, total number of positive integral solution is given by
=(Number of factors of n)/2
Now, x^2  y^2 = 287
287 = 7 * 41
Number of factors = 2 * 2 = 4
Number of positive solution = 4/2 = 2
Taking +/ x and +/ y into consideration
2 * 4 = 8

@zabeer
For equation, x1 + x2 + x3 + ... + xn = s
Number of positive integral solution where s >= n is given by : (s1)C(n1)
So, for a + b + c = 20
= (201)C(31)
= 19c2
= 171

x1 + x2 + x3 + x4 = 20
Total positive solutions of is (20  1)C(4  1) = 19C3 = 969
Let number of positive integral solutions when x1 > x2 = m
Let number of positive integral solutions when x2 > x1 = m
Let number of positive integral solutions when x1 = x2 = nSo, 2m + n = 969
If x1 = x2 = p then 2p + x3 + x4 = 20
p could be 1  9, so 2p would be 2  18
2p = 2, x3 + x4 = 18
2p = 4, x3 + x4 = 16
and so on till 2p = 18, x3 + x4 = 2Number of solutions = 17C1 + 15C1 + 13C1 + ... + 1C1
= 1 + 3 + 5 + ... + 17
= 9^2
= 812m + 81 = 969
m = (969  81)/2 = 444

x^(2/3) + x^(1/3)  42 ≤ 0
Let x^(1/3) = z
z^2 + z  42 ≤ 0
(z + 7)(z  6) ≤ 0
for non negative solutions, 0 ≤ z ≤ 6
z = x^(1/3)
so 0 ≤ x^(1/3) ≤ 6
0 ≤ x ≤ 6^3
0 ≤ x ≤ 216
so we have 217 values from 0 to 216.

@aamin_afzal Perfect!
Just to add, for non negative solutions it would be (s + n  1) C (n  1)

@zabeer When I do 14C4 after adding dummy variable, why I don't get answer.

a + b + c + d ≤ 15
a + b + c + d + e = 15 ( e is a dummy variable and will be zero in case of a + b + c + d = 15)
Now as we have both natural numbers (a, b, c and d) and whole numbers (e) in the same equation, let take e = E  1, where E is natural)
a + b + c + d + E  1 = 15
a + b + c + d + E = 16
number of positive solutions = (n  1) C (k  1)= 15C4 = 1365
(trap here is that when we do a + b + c + ... we should ensure that all of them be either positive or non negative (which includes 0). You can't have both in RHS)