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1
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1
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2
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3
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The rest of the solutions can be derived from that integral solution
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For each successive integral solutions of the equation, the value x and y will
change by a coefficient of the other variable
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If the equation is of the type Ax + By=C,an increase in x will cause a decrease
in y
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2x + 3y = 39
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(Number of Integral Solutions) Step-2: For a given equation, you should start
substituting values (by hit and trial) for the variable that has larger coefficient to find out
first integral solution
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Now, if we take y = 0, we will get x = 39/2(not
an integer)
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So, (18,1) is our first solution
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That means, if we add 2n (where n is an integer) to the first value for y, we
will have to subtract 3n from the first value of x to get integral solutions
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If y= 1 + 2(2) = 5, x= 18 – 3(2) = 12
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(Number of Integral Solutions) Step-4:This equation will have infinite number of
integral solutions but finite number of non-negative integral solutions
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We can keep increasing the value of y in the positive direction but x will be decreasing
simultaneously and become less than 0 at one point
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So, x
can take 7 non negative integral values and they are- 18, 15, 12, 9, 6, 3 and 0
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Note: In equation Ax + By = C, if C is divisible by any of A or B, then number of
non-negative integral solutions = {C/LCM(A,B)} + 1
2
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Let us understand the concept from an example:
X1 + X2 + X3= 8
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8 objects have 7 gaps between them
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These selected gaps will hold the plus signs of the given equation
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Therefore, number of positive integral solutions of equation x1+x2+⋯+xr=n
= Number of ways in which n identical balls can be distributed into r distinct boxes
where each box must contain at least one ball
= (n-1)C(r-1)
Case-2: Number of Non-negative integral solutions

We will continue with our previous equation
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We will substitute the variables in the question such that this case would become similar
to previous case
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In this case, (X1, X2, X3) >= 0
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Substitute X1+1=Y1, X2+1=Y2 and X3+1=Y3 in the
given equation such that
(X1+1) + (X2+1) +(X3+1) = 11
=> Y1+Y2+Y3=11
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Therefore, Number of non-negative integral solutions of equation x1+x2+⋯+xr=n
= Number of ways in which n identical balls can be distributed into r distinct boxes
where one or more boxes can be empty
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-Constraints on the variables
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To solve this, replace A,B,C with P,Q,R such that P= 6-A, Q=6-B and R=6-C
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As A ranges from 1 to 6, P ranges from 0
to 5
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The number of non-negative solutions is 7C2 = 21
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If the linear equation is x1 + x2 +
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… xr ) <=p then the problem can be reduced to finding the exponent of
xnin the expression ( 1 + x + x2 + x3
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3
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Now, for each solution
(x1,y1), there would exist 4 values for x and y, They are-> (x1,y1), (-x1,y1) ,(x1,-y1) and (-x1,y1)
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4
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Case 1: N is an odd number and not a perfect square
In this case, total number of positive integral solutions will be= (Total number of factors
of N) / 2
Example: How many positive integral solutions are possible for the equation X 2– Y2=
135?
Solution: Total number of factors of 135 is 8
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Case 2: N is an odd number and a perfect square
In this case, total number of positive integral solutions will be = [(Total number of
factors of N) – 1] / 2
Example: How many positive integral solutions are possible for the equation X 2– Y2=
121?
Solution: Total number of factors of 121 is 3
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In this case, total number of positive integral solutions will be = [Total number of factors
of (N/4)] / 2
Example: How many positive integral solutions are possible for the equation X2– Y2=
160?
Solution: Total number of factors of 40 is 8 (as N=160 and N/4=40)
So, total number of positive integral solutions = 8/2 = 4
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Case 4: N is an even number and a perfect square
In this case, total number of positive integral solutions will be ={[Total number of factors
of (N/4)] – 1 } / 2
Example: How many positive integral solutions are possible for the equation X 2– Y2=
256?
Solution: Total number of factors of 64 is 7
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Number of Integral Example 1: Find the number of positive integral solutions of |x| +
|y| = 10
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First find the positive integral
solution of a+b = 10
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Now for each solution (a1, b1), the
values of (x,y)= (a1, b1), (-a1, b1), (a1, -b1) and (-a1, -b1)
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Number of Integral Example 2: Find the number of positive integral for a,b,c and d
such that their sum is not more than 15
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a + b + c + d = 14,13,12,11,10,9,8,7,6,5,4
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Number of Integral Example 3: Find the total number of integral solutions ofIxI + IyI +
IzI = 15
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Now, we need to find the
number of positive integral solutions of a + b + c = 15
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Now for each value of a,b and c we will have two values of x, y and z each
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Now let one of the variables be equal to 0
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Therefore, we need the positive integral solution of b + c = 15, where b
= |y| and c = |z|
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Each of these solutions will give
two values of y and z and there are 3 ways in which we can keep one of the variables
equal to 0
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Now let two of the variables be equal to 0
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Therefore, the total number of integral solutions = 728 + 168 + 6 = 902

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