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

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
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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:

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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

similar question for practice

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

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

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

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

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

is zero a positive integer ?

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

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.

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

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.

Great explanation. Thanks Brent BrentGMATPrepNow.

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