If x and y are positive integers, what is the value of x?

Question

1) 3x + 2y = 12
2) |9x2 + 30xy + 25y2| = 212

*kudos for all correct solutions

chetan2u · Accepted Answer

Hi,
A good Q, easy to error..

Catch is x and y are POSITIVE integers

Let's see the statements...
1) 3x + 2y = 12
2y and 12 are even, so 3x will also be even..
Thus x is even... It can be 2 or 4..
Only x as 2 gives us an integer value of y ..
X=2 and y=3..
Sufficient..

2) |9x² + 30xy + 25y²| = 21²
 |(3x+5y)^2=21^2.....3x+5y=21.. ..
Again ONLY x as 2 & y as 3 satisfy the equation..
Sufficient

D

BrentGMATPrepNow · Answer

IMPORTANT: I created this question to highlight a  common myth  about Data Sufficiency questions as well as highlight a common mistake that students make.

Target question :    What is the value of x?

Given : x and y are  positive integers

Statement 1 : 3x + 2y = 12   
Some students will see this equation with 2 variables and automatically conclude that there are infinitely many solutions, in which case, statement 1 is not sufficient. 
Under most conditions, this conclusion would be correct. However, in this question, we have the given condition that x and y are  positive integers , which severely limits the possible solutions. 
In fact, there is only ONE pair of positive integers that satisfy the equation: x = 2 and y = 3. 
So, we can be certain that  x = 2 
Since we can answer the  target question  with certainty, statement 1 is SUFFICIENT

Statement 2 : |9x² + 30xy + 25y²| = 21²  
First factor the part inside the absolute value to get: |(3x + 5y)²| = 21²
This means that EITHER (3x + 5y)² = 21² OR (3x + 5y)² = -(21²)
We can quickly dismiss the second case, (3x + 5y)² = -(21²), since (3x + 5y)² must be greater than or equal to zero. So, it could never equal -(21²)

So, what about (3x + 5y)² = 21²? 
This means that either 3x + 5y = 21 or 3x + 5y = -21 
If x and y are both positive, we know that 3x + 5y will be positive, which means there are no solutions to the equation 3x + 5y = -21 
What about the equation 3x + 5y = 21? 
Under the restriction that x and y are POSITIVE INTEGERS, there is only ONE pair of positive integers that satisfy the equation: x = 2 and y = 3. 
So, once again, we can be certain that  x = 2 
Since we can answer the  target question  with certainty, statement 2 is SUFFICIENT

Answer:  D

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achuepnlx · Answer

Option D.
1) sufficient
3x+2y = 12
x=2 and y=3

2) sufficient
we can reduce the eq to = (3x +5y)^2 = 21^2; 3x+5y = 21
x=2 and y=3

rohit8865 · Answer

similar question for practice

https://gmatclub.com/forum/peter-went-to-the-store-to-buy-paint-small-cans-cost-164747.html

attari92 · Answer

x and y are positive integer
statement 1st 3x+2y =12 , considering the question stem, it will result in only one solution i.e(2,3)
statement 2nd will water down to 3x +5y =21 , considering question stem, it will result in only one solution (2,3)-> option D

hudhudaa · Answer

hi Brent - thanks for posting this question and providing the explanation. when working with absolute values, I thought we would have to consider all possibilities as follows - please help me understand why this approach is incorrect (and since I ended up with 4 possibilities as shown below - I concluded that this statement was insufficient). I can see why case 2 and case 3 are invalid as RHS and LHS are not equal - but what about cases 1 and 4?:

case 1: 3x + 5y = 21

case 2: -(3x + 5y) = 21

and since we took the square root of both sides we were left with + / - 21 on the RHS, so we have two more cases:

case 3: 3x + 5y = -21

case 4: -(3x + 5y) = -21

case 1: 3x + 5y = 21

case 2: -(3x + 5y) = 21

and since we took the square root of both sides we were left with + / - 21 on the RHS, so we have two more cases:

case 3: 3x + 5y = -21

case 4: -(3x + 5y) = -21

amanvermagmat · Answer

Hi

You are right that when we are given (3x+5y)^2 = 21^2 then we have to consider 4 possibilities.

But here in this question we are also given that x/y are positive integers, so 3x+5y has to be a positive integer only, and it cannot be anything else. So we just go with the case 3x+5y = 21, and then we try for positive integer values of x/y (not even 0) which will satisfy the equation. We will see that the only possibility is x=2, y=3

BrentGMATPrepNow · Answer

Good question.

You'll find that the solution to the case 1 equation is identical to the case 4 equation, and the solution to the case 2 equation is identical to the case 3 equation. The reason for this is that equations 1 and 4 are equivalent equations, and equations 2 and 3 are equivalent equations.

Take, for example, case 1: 3x + 5y = 21
If we multiply both sides of this equation by -1, we get -(3x + 5y) = -21 (the case 4 equation).

The same holds true for equations 2 and 3

Cheers,
Brent

anmolsd1995 · Answer

is zero a positive integer ?

BrentGMATPrepNow · Answer

Zero is an integer. 
Zero is neither positive nor negative.

GMATinsight · Answer

anmolsd1995 Zero is 1) Neither Positive, nor a Negative integer but Yes it's an integer 2) Zero is an Even integer 3) Zero is a special highlight of Non-negative integers over positive integers.

Kimberly77 · Answer

Hi Brent  BrentGMATPrepNow , at this stage since we are squaring both sides, so shouldn't it be just 3x + 5y = 21 only? Have I missed something here?
(3x + 5y)² = 21²? 
This means that either 3x + 5y = 21 or 3x + 5y = -21

BrentGMATPrepNow · Answer

Here's an example to show why that reasoning is incorrect: 
We know that 3² = (-3)²,  since both sides of the equation evaluate to be 9 
However, we can't just remove the exponents and conclude that 3 = -3

Similarly, if x² = 9, we can't conclude x = 3 is only one solution, since x = -3 also satisfies the equation.

Kimberly77 · Answer

Great explanation. Thanks Brent  BrentGMATPrepNow .

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If x and y are positive integers, what is the value of x?

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