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

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by Brent@GMATPrepNow » Thu Sep 12, 2019 10:58 am
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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

Source: www.gmatprepnow.com
Difficulty level: 650-700
Answer: D
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by Brent@GMATPrepNow » Sat Sep 14, 2019 3:21 pm
Brent@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

Source: www.gmatprepnow.com
Difficulty level: 650-700
Answer: D
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 11
From this, we can conclude that 4(2 – y) is a multiple of 11

What 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 = 17
So, the LEAST possible value of x+y = 17 + 13 = 30

Answer: D

Cheers,
Brent
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by deloitte247 » Thu Sep 19, 2019 9:32 am
$$15y-11x=8$$
$$15y=8+11x$$
Therefore, (8+11x) must be divisible by 15 which is coefficient of y since 15 is divisible by 5 and 3 the (8+11x) must be divisible by both 5 and 3.
For a number to be divisible by 5, the number must end in either 5 or 0
Therefore, (8+11x) must produce a number that ends in 5 or 0 or both x and y are greater than zero. Both x and y are greater than 3. i will plug in some values of x
If x = 5
$$\left[8+11\left(5\right)\right]=8+55=63$$\
This is not divisible by 5, so $$x\ne5$$
Even though 63 is divisible by 3

If x=12
$$\left[8+\left(11\cdot12\right)\right]=8+132=140$$
This is divisible by 5, but not divisible by 3, so $$x\ne12$$

If x = 17
$$\left[8+\left(x\cdot17\right)\right]=8+187=195$$
This is divisible by both 5 and 3 so, x =17

$$15y-11\left(7\right)=8$$ $$15y-187=8$$ $$15y=8+187$$ $$\frac{15y}{15}=\frac{195}{15}$$ $$y=13$$ $$So,\ y=13,\ x=17$$ $$x+y=17+13=30$$

$$Answer\ is\ Option\ D$$
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