Question

For the given pair (x,y) of positive integers 3x-11y=1 and given that x can have maximum value of 1000,how many pairs are possible

Answer 1

Answer:

Hey mate plzz refer to the attachment

Answer 2

Answer:

91

Step-by-step explanation:

The given relation is 3x-11y=1, where x and y both are positive integers.

Now, we have to find the number of pairs of (x,y) where x can have maximum value of 1000.

Here, the relation can be written as $y=\frac{3x-1}{11}$ ...... (1)

For y to be an integer (3x-1) has to be a multiple of 11.

Assume that, the values of (3x-1) are 11, 22, 33, 44, 55, 66, 77, 88, 99, 110, ... so on.

Now, for (3x-1)=11 we have, x=4 and y=1.

For (3x-1)= 22 we have, x is fraction.

For (3x-1)= 33 we have, x is fraction.

For (3x-1)= 44 we have, x=15 and y=4

For (3x-1)= 55 we have, x is fraction.

For (3x-1)= 66 we have, x is fraction.

For (3x-1)= 77 we have, x=26 and y=7.

For (3x-1)= 88 we have, x is fraction.

For (3x-1)= 99 we have, x is fraction.

For (3x-1)= 110 we have, x=37 and y=10.

And so on.

Now, the values of x are 4,15,26,37, .... up to nth term, where nth term<1000. This is an A.P. series with first term 4 and common difference 11.

Assume, 4+(n-1)11=1000, ⇒(n-1)11=996, ⇒n-1 =90.54, ⇒n= 91.54

Now, n must be positive integer and also the nth term <1000.

Therefore, n<91.54, i.e. n=91

So, there can be 91 pairs of (x,y) values. (Answer)