Question

If x>y are positive integers such that 3x+11y leaves remainder 2 when divided by 7 and 9x+5y leaves a remainder 3 when divided by 7, then find the remainder when x-y is divided by 7.

Answer 1

Given :

The two given  positive integers x and y are  such  that :

x > y

Remainder  obtained when 3x + 11y is divided by 7 = 2

Remainder obtained when 9x + 5y is divided by 7 = 3

To Find :

The remainder obtained when x - y is divided by 7 = ?

Solution :

∴ It is given that 3x + 11y leaves a remainder 2 when divide by 7 , So by using Euclid's Division Algorithm we can write :

3x + 11y = 7k + 2     -(1)

Similarly we can also write :

9x + 5y = 7k' + 3    -(2)

Now subtracting eq (1) from eq (2) we get :

6x - 6y = 7(k' - k) + 1

Or , 6(x-y) = 7q + 1

Here we have taken q = k' k

And also let x - y = c

so , 6c= 7q + 1

Or, 6c - 1 = 7q

i.e. $\frac{7}{6c} = 1$

Or ,$6c \equiv1 (mod 7 )$

Or , $6 \times 6^{-1} \times c \equiv 1 \times 6^{-1} (mod 7)$

Or , $1 \times c \equiv 6 ( mod 7 )$

For finding $6^{-1}$ :

$(6 \times k ) = 1$  , where k = inverse of 6

⇒ k = 6

∴$c \equiv 6$  (mod 7 )

⇒$\frac{7}{c}- 6$

⇒$\frac{7}{(x - y) } - 6$

i.e. the remainder when x-y id divided by 7 is 6

So ,when x - y is divided by 7 , the remainder is 6.