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
Given: 15y – 11x = 8
Subtract 4y from both sides to get: 11y – 11x = 8 – 4y
ASIDE: Why did I subtract 4y from both sides?
This allows me to factor the expression 11y – 11x, which may help reveal some useful relationship.
Continuing along….
Factor both sides to get: 11(y – x) = 4(2 – y)
KEY CONCEPT: Since x and y are INTEGERS, we know that (y – x) is an INTEGER, which means 11(y – x) is a
multiple of 11From this, we can conclude that 4(2 – y) is a
multiple of 11What is the smallest value of y (given that y is an integer greater than 3) such that 4(2 – y) is a multiple of 11?
If y=
13, then 4(2 – y) = 4(2 – 13) = 4(-11) = -44. Perfect!
So, y=
13 is the smallest value of y to meet the given conditions.
To find the corresponding value of x, take 15y – 11x = 8 and plug in y=13 to get: 15(13) – 11x = 8
Simplify : 195 – 11x = 8
Subtract 195 from both sides: -11x = -187
Solve: x =
17So, the LEAST possible value of x+y =
17 +
13 = 30
Answer: D
Cheers,
Brent
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