If x and y are integers greater than 3, and 15y – 11x = 8,

15y-11x=8

15y=11x+8

8=8 mod 15
Hence, 11x=7 mod 15
-4*x=-8
x=2
At x=2, y=2

x,y>3
As slope of 15y=11x+8 is 11/15, there is 11 units increment in y co-ordinates for every 15 increment in x co-ordinates.
Next integral solution is x=2+15=17 and y=2+11=13
x+y=17+13=30

While all the above methods are to the point and clear, in a case you are not able to think about these methods then this question can be solved backwards from the answer choices by using the method of solving a system of two linear equations and looking for a solution in which y is an integer as x and y are integers as per question stem.

Eq-1: \(15y-11x=8\)
Eq-2: \(x+y=\) Any of the answer options

In case of D,

Eq-1: \(15y-11x=8\)
Eq-2: \(x+y=30\)

Multiplying the second equation with 11 we get - \(11x+11y=330\)
Solving both the equations we get \(26y=338\) -> \(y=13\) which is an integer and satisfies the condition provided in the question stem.

Ans. D

15y=8+11x

y=(8+11x)/15

now x+y= (8+11x)/15 + x

which is equal to

(8+26x)/15

So the answer should be a multiple of 15 .

OptionD

We have: 15y – 11x = 8

Since the difference between 15y and 11x is a constant, we can keep increasing the values of x and y in such a way that their difference remains constant at 8

However, we need to find the minimum value of x + y where x and y are integers greater than 3 (i.e. definitely positive quantities!)

We have: 15y = 8 + 11x => y = (8 + 11x)/15

The smallest possible positive integer value of x so that y remains an integer is clearly x = 2 ---> then y = 2

However, we need values of x and y that are greater than 3

Observe: 15y - 11x = 8

Let me increase y by a certain quantity q and x by quantity p so that the difference remains 8

=> 15(y + q) - 11(x + p) = 8

=> 15y - 11x + (15q - 11p) = 8

=> 15q - 11p = 0

=> p : q = 15 : 11

Since 15 and 11 have no common factors, and since p and q must be integers (since x and y are integers), we must have p = 15 and q = 11

Note that we had a solution x = 2 and y = 2

Thus, the next solution is: x = 2 + p = 17 and y = 2 + q = 13

=> Minimum value of (x + y) = 17 + 13 = 30

Answer D

Highlighted(Red) portion in your solution is not correct.
x and y can be both even or both odd.

(2,2) is a solution of the equation.

\(15y – 11x = 8\)

Or, \(15y = 11x + 8\) , now here RHS must be divisible by 15 = RHS must be divisible by both 3 & 5

Now, try pugging in values for x > 3

For the number to be divisible by 5 the number must end either in 5 or 0

Now , try x = 12 first as units digit will end in 0

11*12 + 8 = 140 , not divisible by 3

Next try x = 17 , as units digit will end in

11*17 + 8 = 195 , Divisible by both 3 and 5

So, we get x = 17 and y = 13

Thus, x + y = 17 + 13 = 30, Answer must be (D)

We can see that 15 - 11 = 4 --> 15y - 11x = 8 is equivalent to 15y - 11x = 2(15 - 11)
--> 15y - 11x = 2.15 - 2.11
--> 15(y - 2) = 11(x-2)
Now, because the question asks "what is the least possible value of (x+y)?" and because the LCM of 15 and 11 is 15.11 --> y - 2 = 11 and x - 2 = 15 --> y = 13 and x = 17. --> (x+y) = 30 (D)

15y-11x=8
15y=8+11x
15(x+y)=8+26x
So (x+y)=(8+26x)/15=2(4+13x)/15

For (x+y) to be minimum, x must be minimum.
x>3
Also for (4+13x) to be divisible by 15(since x,y both are integers, x+y is an integer) x can take values such as 2,17,...
Since x>3, x=2 is out.

For x=17,

X+y = 30

Posted from my mobile device

Given:
1. x and y are integers greater than 3
2. 15y – 11x = 8

Asked: What is the least possible value of x+y?

15y-11x = 8
15y = 11x + 8
x =2; y = 2; 30=22+8; Is a solution but x,y>3
y=13;x=17; 15*13 = 195 = 187+8

x + y = 13 + 17 = 30

IMO D

nick1816 - Could you please explain to me arithmetic modulus part? I am not familiar with the concept but I have heard of it. Where can I read this from?

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