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CAT 2025 Lesson : Linear Equations - Integer Solutions

4.6 Integer solutions

These questions have one linear equation with 2 variables. We need to find the possible number of integral solutions that satisfy. These questions are not standard linear equations questions.

Type 1: In these we can represent one variable in terms of another such that the other variable is in the denominator and a constant in the numerator, and then find the values that satisfy. There will be a finite set of integral solutions in these questions.

Example 12

Where

x

and

y

are positive integers, how many solutions exist for

xy + 2x = 24

Solution

xy + 2x = 24

⇒

xy = 24 - 2x

⇒

y = \dfrac{24}{x} - 2

To get integral solutions,

x

should perfectly divide

24

. Therefore,

x

is a factor of

24

.

Number of positive factors of

24

2^{3} \times 3 = (3 + 1) \times (1 + 1) = 8

However, the first

2

factors, i.e.

x = 1

and

x = 2

will result in a negative value of

y

\therefore

Number of positive integral solutions

= 8 - 2 = \bm{6}

Answer:

6

solutions

Example 13

How many pairs of positive integers

(m, n)

satisfy

\dfrac{1}{m} + \dfrac{4}{n} = \dfrac{1}{12}

where

n

is an odd integer less than

60

?

(

1

)

6

(

2

)

4

(

3

)

7

(

4

)

5

(

5

)

3

Solution

\bm{n}

should be an odd positive integer less than

\bm{60}

, we shall express

\bm{m}

in terms of

\bm{n}

and substitute.

\dfrac{1}{m} + \dfrac{4}{n} = \dfrac{1}{12}

⇒

\dfrac{1}{m} = \dfrac{1}{12} - \dfrac{4}{n}

⇒

\dfrac{1}{m} = \dfrac{n - 48}{12n}

⇒

m = \dfrac{12n}{n - 48}

For

\bm{m}

to be positive,

\bm{n}

can take values of

49, 51, 53, 55, 57

and

59

12n

is divisible by

(n - 48)

only for

\bm{3}

values of

n

, i.e.,

49, 51

and

57

.

Answer: (

5

)

3

Type 2: When the equation is of the form

ax + by = c

, where

a

b

and

c

are constants, and

x

and

y

are given to be integers, then when one variable is written in terms of the other, the other variable is in the numerator. Therefore, there are infinite set of integral solutions here.

In these questions, we should
1) Simplify the equation to a form where the coefficients

a

and

b

are co-prime.
2) Find the smallest integral value of

x

where

y

is also an integer. (Apply the Chinese Remainder concept)
3) The next solution will be where

x

is increased by

b

(the coefficient of

y

) and

y

is decreased by

a

and so on.
4) Continue to do this till the values are within the range (in this case positive).

Example 14

Where

x

and

y

are positive integers, how many solutions exist for

9x + 15y = 291

Solution

We can begin by simplifying the equation as it is divisible by

3

9x + 15y = 291

⇒

3x + 5y = 97

⇒

y = \dfrac{97 - 3x}{5}

x = 4

is the smallest

x

-value where

(97 - 3x)

is divisible by

5

x = 4

y = 17

.

As the sum is constant in the equation, an increase in

x

will result in a decrease in

y

. And,

x

-values will increase by the coefficient of

y

, i.e.

5

and

y

-values will decrease by the coefficient of

x

, i.e.

3

.

Next solution ⇒

x = 4 + 5 = 9, y = 17 - 3 = 14

Next solution ⇒

x = 9 + 5 = 14, y = 14 - 3 = 11

This process continues till

y

remains positive. Using this movement of

x

and

y

values, we get the following

\bm{6}

solutions for

(x, y)

⇒

(4, 17); (9, 14); (14, 11); (19, 8); (24, 5); (29, 2)

Answer:

6

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