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

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Joined: 12 Sep 2015
Posts: 4225
If x and y are integers greater than 3, and 15y – 11x = 8,  [#permalink]

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10 Sep 2019, 13:25
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65% (hard)

Question Stats:

59% (02:41) correct 41% (03:15) wrong based on 69 sessions

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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

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If x and y are integers greater than 3, and 15y – 11x = 8,  [#permalink]

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10 Sep 2019, 14:25
1
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

Let's analyze the equation

15y – 11x = 8 ....we can find simple observation

Odd -Odd = Even....So x & y must be odd to maintain even number (8).

Let's alter the equation:

15y = 11x + 8

x must be have unit digit of 7 or 2 to make the sum equal either to 5 or 0 to divide 15 but as mentioned it x can't be even so focus on 7,17,...etc

Let x=7.....15y = 85...not divisible by 15 (as we need number to be divisible by 3 too)

Let x=17....15y = 195....bingo...We can Eliminate A & B directly.

Use choice C that represented x+y......26-17 =9 but 9 * 15 does not equal to 195

Use choice D that represented x+y......30-17 =13..... 13 * 15 equal to 195

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Joined: 19 Oct 2018
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If x and y are integers greater than 3, and 15y – 11x = 8,  [#permalink]

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Updated on: 11 Sep 2019, 05:18
1
1
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

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

Originally posted by nick1816 on 11 Sep 2019, 02:07.
Last edited by nick1816 on 11 Sep 2019, 05:18, edited 1 time in total.
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Re: If x and y are integers greater than 3, and 15y – 11x = 8,  [#permalink]

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11 Sep 2019, 02:12
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.

Mo2men wrote:
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

Let's analyze the equation

15y – 11x = 8 ....we can find simple observation

Odd -Odd = Even....So x & y must be odd to maintain even number (8).

Let's alter the equation:

15y = 11x + 8

x must be have unit digit of 7 or 2 to make the sum equal either to 5 or 0 to divide 15 but as mentioned it x can't be even so focus on 7,17,...etc

Let x=7.....15y = 85...not divisible by 15 (as we need number to be divisible by 3 too)

Let x=17....15y = 195....bingo...We can Eliminate A & B directly.

Use choice C that represented x+y......26-17 =9 but 9 * 15 does not equal to 195

Use choice D that represented x+y......30-17 =13..... 13 * 15 equal to 195

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Joined: 12 Sep 2015
Posts: 4225
Re: If x and y are integers greater than 3, and 15y – 11x = 8,  [#permalink]

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11 Sep 2019, 05:45
3
Top Contributor
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 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

Cheers,
Brent
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Re: If x and y are integers greater than 3, and 15y – 11x = 8,  [#permalink]

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11 Sep 2019, 08:38
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

$$15y – 11x = 8$$

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)
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Re: If x and y are integers greater than 3, and 15y – 11x = 8,  [#permalink]

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11 Sep 2019, 08:50
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:
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
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If x and y are integers greater than 3, and 15y – 11x = 8,  [#permalink]

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29 Sep 2019, 00:51
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
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Re: If x and y are integers greater than 3, and 15y – 11x = 8,  [#permalink]

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30 Sep 2019, 05:37
nick1816 wrote:
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

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

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?
Re: If x and y are integers greater than 3, and 15y – 11x = 8,   [#permalink] 30 Sep 2019, 05:37
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