GMATPrepNow wrote:
If x and y are integers greater than 3, and 15y – 11x = 8, what is the least possible value of x+y?
A) 12
B) 16
C) 26
D) 30
E) 34
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