If x and y are positive integers such that x^2 – y^2 = 48, how many

Question

If x and y are positive integers such that  x^2 – y^2 = 48 , how many different values of  y  are possible?

(A) Two

(B) Three

(C) Four

(D) Five

(E) Six

CrackverbalGMAT · Answer

x and y are positive and  x^2 - y^2 = (x + y) \space (x - y) = 48

(i) Now x + y has to be positive and to get a positive value 48, x - y also has to be positive. Therefore x > y and x, y  
eq  0

(ii) 48 is an even number. Therefore  x^2 \space and \space y^2  also have to be even (E - E = E)

So x and y are also even

Different ways of getting 48 as as a product of 2 even numbers

(a) 1 * 48 => Not possible as 1 number is even and the other is odd
 
(b) 2 * 24 => Possible. x = 13 and y = 11

(c) 3 * 16 => Not possible as 1 number is even and the other is odd
 
(d) 4 * 12 => Possible. x = 8 and y = 4

(e) 6 * 8 => Possible. x = 7 and y = 1

Therefore there are 3 possible values of y

Option B

Arun Kumar

TestPrepUnlimited · Answer

(x - y)(x + y) = 48 . We need integer solutions so break down 48 into 48 = 1*48 = 2*24 = 3*16 = 4*12 = 6*8. (x - y) is the first number in each pair while (x+y) is the 2nd number. However not all of these combinations are viable, the sum of the two factors in any pair is  x - y + x + y = 2x , so we need the sum of the factors to be even.

1*48 is not acceptable, 2*24 is good. 3*16 makes 2x = 19 so we cannot take that, but we can take 4+12 and 6+8. Therefore there are 3 options viable.

Ans: B

arvind910619 · Answer

Dear Sir,

I did not understand the highlighted part. 
Odd- Odd = Even
How did you deduce that both the number have to be even?

ScottTargetTestPrep · Answer

Solution:
 
Since x^2 - y^2 = (x - y)(x + y), we have (x - y)(x + y) = 48. Since 48 = 1 * 48 = 2 * 24 = 3 * 16 = 4 * 12 = 6 * 8, and x - y < x + y, we have the following possible systems of linear equations:

1) x - y = 1 and x + y = 48

2) x - y = 2 and x + y = 24

3) x - y = 3 and x + y = 16

4) x - y = 4 and x + y = 12

5) x - y = 6 and x + y = 8

However, since both x and y are integers, if we add the two equations in each system, the right hand side of the resulting equation can’t be an odd integer. For example, if we take the first system, and add the two equation, we have:

2x = 49

x = 49/2

We see that x is not an integer. Therefore, this eliminates the first system. It also eliminates the third system using the same argument. However, the other three systems are fine since they yield integer values for x and y. Therefore, there are three different values of y possible.

Answer: B

Fdambro294 · Answer

A Positive Integer can be written as a Difference of Two Perfect Squares only when:

-X and Y are both Even Integers

-X and Y are both Odd Integers

In order to have 2 positive integer Perfect Squares subtract to another positive Integer ———-> The Factors of: (X + Y) (X - Y) must be equal to Values such that when you look on the number Line, the Value of X is the Mid Point between (X - Y) and (X + Y) and this X Midpoint must be a Positive Integer

Stated more succinctly, the Mid-Point between (X + Y) and (X - Y) must be a Positive Integer

Listing out the Factor Tree of 48, which are the possibilities we can have the Factors of (X + Y) and (X - Y) be in order to multiply to 48:

1————48

2————-24

3————16

4————-12

6————8

The Mid-point of any 2 Values such as (X - Y) and (X + Y) = Arithmetic Mean of those two Values = (Value 1 + Value 2) / (2)

Thus, as stated above, the values picked must both be Odd or both be Even.

Only 3 Factor Pairs satisfy this requirement:

(2 , 24)

(4 , 12)

(6 , 8)

If you understand the underlying rationale for why this must be true, you can pick 3 and move forward. Otherwise you can prove why it works for each factor pair and find the individual Values of X and Y

Case 1: (2 , 24)

It will always be True that, for any 2 given positive integers, the Value of (X + Y) will always exceed the Value or (X - Y)

Thus:

(X - Y) = 2

(X + Y) = 24

Add the two equations and you are left with:

2X = 26 ———> X = 13

And Y = 11

Proof:

(X + Y) (X - Y) = (X)^2 - (Y)^2 = (13)^2 - (11)^ 2 = 169 - 121 = 48

You can do the same thing for the other 2 cases:

Case 2: factor pair of (4 , 12) ————> X = 8 and Y = 4 ———-> 64 - 16 = 48

Case 3: factor pair of (6 , 8) ————-> X =
7 and Y = 1 ———-> 49 - 1= 48

Answer: three

Posted from my mobile device

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

If x and y are positive integers such that x^2 – y^2 = 48, how many

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

If x and y are positive integers such that x^2 – y^2 = 48, how many

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards