Solved Questions On Finding Number Of Integral Solutions Concept  Vikas Saini

Q.1) How many integral values of x satisfy the equation x = 2x  1203x ?
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.Q. 2) How many integral values of (x,y) satisfy the equation x^2 – y^2 = 627 ?
x^2 – y^2 = 627.
627 = 3 x 11 x 19.
Total number of factors = 8.
This can be done in negative too.
Total integral values = 2 x 8 = 16.Q.3) Find the number of positive integral positive solution for a+b+c = 20.
By direct formula,
x1 + x2 ... xK = N.
Positive solution = (N – 1)C(K1)
= (20 – 1)C(31)
= 19C2
= 171.Q.4) Find the number of positive integral solutions of x1+x2+x3+x4 = 20, where x1 > x2 ?
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.Q.5) Find the number of positive integral for a,b,c and d such that their sum is not more than 15.
a + b + c + d < 15.
a + b + c + d = 14,13,12,11,10,9,8,7,6,5,4.
Total no of positive solution = 13C3 + 12C3 ... 1
= 286+220+165+120+84+56+35+20+10+4+1
=1001.Q.6) How many integral values of x wil satisfy x3+2x+4+x < = 11.
X = 0,1,2,1,2,3.
Total solution = 6 values.Q.7) Number of integral solutions of x^2 – y^2 = 287 ?
287 = 7 x 41.
Total no of factors = 2 x 2 = 4.
It can be written in negative form also.
Total integral solution = 2 x 4 = 8.Q.8) Total number of integral solution of a^3 + b^3 + c^3 = 43655.
Not any number can be written as 9n+4.
43655 mod 9 = 4 or 5.
Hence 0.Q.9) Find the total no of integral solutions of w^4 + x^4 + y^4 + z^4 = 1797.
Unit digit of N^4 = 1,6,5, where N is a natural number.
Unit digit of Sum of w^4,x^4,y^4,z^4 never can be get as 7.
Hence 0 solution.Q.10) How many integral values of x will satisfy x^2 – 2x8 > = 0.
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.Q.11) Find the number of total integral solutions of x^2 + y^2 = 100.
x^2 + y^2 = 100.
(0,10)(10,0)(10,0)(0,10),(6,8)(6,8),(6,8),(6,8)
Total solution =8.Q.12) Number of integral solution of x^2 + y^2 = 36.
x^2 + y^2 = 36.
(0,6)(0,6),(6,0),(6,0)
Total = 4 solutions.Q.13) Number of integral solutions x^2 + y^2 < 36.
Sum of square possible below 36 are 0,1,4,9,16,25,2,10,17,26,8,13,20,29,18,34,32.
x^2 + y^2 = 0 (0,0)
1 solution
X^2 + y^2 = 1[(0,1),(0,1),(1,0),(1,0)]
4 solution.
X^2 + y^2 = 9. [(0,3)(0,3),(3,0)(3,0)]
4 solution.
X^2 + y^2 = 16. [(0,4),(0,4),(4,0)(4,0)
4 solutions.
X^2 + y^2 = 25. [(0,5),(0,5),(5,0),(5,0),(3,4),(3,4),(3,4),(3,4)
8 solutions.
As same as for sum = 2,8,18,32.
Each has 2 solutions.
And sum = 10,17,26,13,20,29,34.
Each has 4 solutions.
Total solutions = 1+4+4+4+8+2x4+4x7
= 57.Q.14) Number of integral solutions of equation a+b+c+d = 30.
Where a >=2, b>= 0, c>=5, d >=8.(a+2)+(b+0)+(c5)+(d+8) = 30.
a+b+c+d = 25.
Total integral solutions = (25+41)C(41) = 28C3 = 3276.Q. 15) Find number of positive integral solution of abc = 120 ?
abc = 120.
120 = 2^3 x 3 x 5.
By direct formula,if N = x^a x y^b x z^c
Then total solution = (a+2)C2 x (b+2)C2 x (c+2)C2
Total number of solution = (3+2)C2 x (1+2)C2 c (1+2)C2
= 10 x 3 x 3
= 90.Q.16) If E = x + x – 1 + x2 + x3 + x4, for how many integral values of x is E less than or equal to 54 ?
x = 8 to 12.
Total 21 values.Q.17) Find number of positive integral values satisfying (x^2 + 5x + 6 )^(x1) = 36.
36 = 6^2, 3^2 x 2^2.
x– 1 = 2.
X = 3.
X^2 + 5x +6 = (x+3) (x+2)
(x+3)^(x1) (x+2)^(x1) = 3^2 x 2^2.
x= 0.
None of the value of x satisfy here.
Hence 0.Q.18) Find number of positive integral values satisfying (x+y)^(y1) = 100.
100 = 10^2 , (10)^2.
Y  1 =2.
Y = 3.
(x + y) =10, 10.
x + 3 = 10, 10.
x = 7, 13.
Hence two solution.Q. 19) Determine the number of positive integral solutions (x,y) to x^y = y^90.
90 = 2 x 3^2 x 5.
No of factors =2 x 3 x 2 = 12.
Factors are 1,2,3,5,6,9,10,15,18,30,45,90.
(x,y)=(1,1),(2^89,2),(3^87,3),(5^85, 5),(6^84,6),(9^81,9),(10^80,10),(15^75,15),(18^72,18),(30^60,30),(45^45,45),(90,90)
Hence 12 positive solutionQ.20) Determine the number of positive integral solutions (x,y) to x^y = y^100.
100 = 2^2 x 5^2.
No of factors = 3 x 3 = 9.
Factors = 1,2,4,5,10,20,25,50,100.
(x,y) = (1,1),(2^98,2),(4^96, 4),(5^95,5),(10^90,10),(20^80,20),(25^75,25),(50^50,50),(100,100).Hence 9 positive solution.
Q.21) Determine the number of positive integral solutions (x,y) to x^y = y^72.
72 = 2^3 x 3^2.
No of factors = 4 x 3 = 12.
Factors = 1,2,3,4,6,8,9,12,18,24,36,72.
(x,y) = (1,1),(2^70,2), (3^69,3)(4^68,4),(6^66,6),(8^64,8),(9^63,9),(12^60,12),(18^54,18),(24^48,24),(36^36),(72,72).
Hence 12 solutions.Q.22) How many pairs of integers (a,b) are possible such that a^2 – b^2 = 288 ?
288 = 2^5 x 3^2.
No of factors = 6 x 3 = 18.
Total no of integer solution = 18.Q.23)Find unordered number of integral solutions of a x b x c = 110.
110 = 2 x 5 x 11.
Factors = 8
No of solution = 3 x 3 x 3 = 27.
No of integral solution = 4 x 27 = 108.
Unordered integral solutions = 108 / 3! = 18.
Hence 18 solutions.Q.24) Find the integral solutions of x + y = 7.
Direct formula
x + y = n.
Total solution = 4n.
Total integral solution = 4 x 7 = 28.Q.25)Find integral solution of x + y < 7.
x + y = 0,1,2,3,4,5,6.
x+y = 0.
Solution : 1.
x + y = 1.
As per direct formula, total solution = 4 x 1 = 4.
Total solutions = 1 + 4(1+2+3+4+5+6)
Total solution = 85.Q.26) Find integral solutions x + y = < 7.
x+y = 0,1,2,3,4,5,6,7.
Total solutions = 1 + 4(1+2+3+4+5+6+7)
= 113.Q.27) Find the integral solutions of x+2 + y = 7.
By direct formula,
Total solutions = 4 x 7 = 28.Q.28) Find the integral solutions of x+2+y < = 7.
By direct formula,
Total solutions = 1 + 4(1+2+3+4+5+6+7)
= 113.Q.29) Find the integral solutions of x+2 + y+4 = 7.
By direct formula,
Total solution = 4 x 7 = 28.Q.30) Find the integral solutions of x+2 + y+4 < = 7.
By direct formula,
Solutions = 1 + 4(1+2+3+4+5+6+7)
= 113.Q.31) Number of integral solutions xy = 2x – y.
xy = 2x – y.
Divided by xy both sides
1 = 2/y – 1/x.
1 = (y2) (x+1)
2 = (x+1) (y2)
(1) x (2) = (x+1) (y2)
(2) x (1) = (x+1) (y2)
(2) x (1) = (x+1) (y2)
1 x (2) = (x+1) (y2)4 integral solutions
Q.32) Total no of positive integral solutions of a + b + c + d < 14.
Let’s take a dummy variable e
a + b + c + d + e = 14.
By direct formula
Positive solutions = (14 – 1)C(51)
= 13C4
= 715.Q.33)How many integral possible for below given equation 3/x – 2/y= 1/12.
36/x – 24/y = 1.
(x – 36) (y + 24) = 36 x 24.
36 x 24 = 2^2 x 3^2 x 2^3 x 3
= 2^5 x 3^3.
Total solutions = 24.Q.34) How many integral pairs (x,y) satisfy 2^x – 3^y = 55 ?
2^x = 55 + 3^y.
X = 6, y =2.
Only 1 solutionQ.35) Number of integral solutions of 1/x + 1/y = 1/48.
48/x + 48/y = 1.
(x48) (y48) = 48 x 48.
48^2 = 2^8 x 3^2.
Total no of factors = 9 x 3 = 27.
Hence no of integral solutions = 27.Q.36) Number the number of positive integral solutions of the equation a(a^2 – b) = (b^3 + 61)
a^3 – ab = b^3 + 61.
a^3 – b^3 = ab + 61.
a = 6, b = 5.
Only 1 solution.Q.37) How many natural solutions for possible (x+1) (x3) (x+5)(x7)(x+9).........................(x99) < 0.
If we plot above equation on xy axis then on positive side x = 3,7..........99.
On negative side x = 1,5...................97.
Total values = 75.Q.38) Find the number of integral solution of x^2 + y^2 = 70.
Sum of square of two numbers can’t be in form of 4n + 2.
Hence 0 solution.Q.39) How many natural solutions are possible for the equation 2^n – 7 = x^2.
2^n = x^2 + 7.
n = 3 then x = 1.
n = 4 then x =3.
n =5 then x = 5.
n = 7 then x = 11.
and so on.Q.40) If 4y – 3x = 5, what is the smallest integer value of x for which y > 100 ?
4y – 5 = 3x.
y = k + 100.
4(k + 100) – 5 = 3x.
4k + 400 – 5 =3x.
4k + 395 = 3x.
K =1 then x = 133.
Smaller value of x is 133.Q.41) Total number of integral solutions of 13x – 3y = 1000 for 100 < x < 200.
13x = 3y + 1000.
y = 113, then x = 103.
Y = 126, then x = 106.
Y = 139, then x = 109.
So on..
X = 103,106,109..............199.
Total 33 values.

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