a, b, c, and d are positive integers. If (a + b)(c d) = r, where

Question

a, b, c, and d are positive integers. If (a + b)(c – d) = r, where r is an integer, is √(c + d) an integer?

(1) (a + b)(c + d) = r^2
(2) (a + b) = x^4*y^6*z^2, where x, y, and z are distinct prime numbers.

KarishmaB · Accepted Answer

All this question is trying to do is confuse you with a ton of variables. The concept being tested here is simple - the prime factors of a perfect square have even powers.

"is √(c + d) an integer?" is just another way of saying "is (c+d) a perfect square?"

(1)  (a + b) (c + d) = r^2 
 (a + b) (c + d) = (a + b)^2 * (c - d)^2 
 (c+d) = (a+b)(c - d)^2 
We know now that (c+d) is a product of (a+b) and a perfect square. If (a+b) is a perfect square too, then (c+d) is a perfect square. Else it is not. Not sufficient.

(2) (a + b) = x^4 y^6 z^2, where x, y, and z are distinct prime numbers.
This tells us that (a+b) is a perfect square. Though alone it is not sufficient since it doesn't tell us anything about (c+d), together with statement 1, it is sufficient.

Answer (C)

mikemcgarry · Answer

I'm happy to help with this. :-)

Statement #1 : (a + b) (c + d) = r^2 The prompt told us that (a + b) (c – d) = r. If we divide the statement #1 equation by the prompt equation, we get / = r, which is intriguing, but which doesn't, by itself, tell us anything about whether (c + d), the numerator, is a perfect square. This is insufficient . Statement #2 : (a + b) = x^4 y^6 z^2, where x, y, and z are distinct prime numbers This one definitively tells us that (a + b) is a perfect square, so we know a perfect square times (c - d) equals r, but we know nothing at all about (c + d). This one by itself, doesn't tell us anything. This is insufficient . Now, consider the statements combined. Statement #1 : (a + b) (c + d) = r^2 Statement #2 : (a + b) = x^4 y^6 z^2, where x, y, and z are distinct prime numbers Now, we know that (a + b) is a perfect square, because it has even powers of all prime factors. We also know that r^2 is a perfect square, so it must also have even powers of all prime factors. This can only mean that (c + d) has even powers of all prime factors, and therefore must be a perfect square. Another way to say that --- we could re-arrange the statement #1 equation to (c + d) = / , and we know this quotient is a positive integer. If the ratio of two squares is an integer, that integer must also be a perfect square. Either way, the combination of statements is now sufficient to give a definitive answer to the prompt. Answer = C One thing that's a little unusual about this question --- with the information in both statements, we could answer the prompt question, but as it turned out, the equation given in the prompt was irrelevant. I don't know that this would happen on the GMAT. Let me know if anyone reading this has any questions. Mike :-)

jlgdr · Answer

OK I will give it a try We are told that (a+b)(c-d) is an integer. We need to find if (c+d) is a perfect square Statement 1 (a+b)(c+d) = Perfect square We also have that (a+b)(c-d) is an integer. Let's say x = a+b Then we have that x(c+d) perfect square and x(c-d) is an integer. What does this tell us? That x = c+d or that a+b = c+d. But not enough to answer the question Statement 2 (a+b) = x^4y^6z^2 Clearly insuff Both together If (a+b)(c+d) is a perfect square and then (a+b) = x^4y^6z^2 then it must follow that (c+d) is AT LEAST X^2*Z^4 to make it a perfect square Is X^2*Z^4 a perfect square? Well we are told that x and z are integers so yes Answer is C Hope it helps Kudos rain!! Cheers! J

jlgdr · Answer

Thanks Karishma, that was precise.

Do you agree with Mike in that it is unlikely that information given in the prompt of a DS question will not be used? Honestly I've never seen a question in which information is given and then not used. Haven't even seen a case such as this one in which it could be used for one of the statements only

Please advice should we worry about this
Thanks!
Cheers
J

KarishmaB · Answer

I agree that the question is not very well thought out. It has a lot of clutter as if the question maker had something else in mind but it didn't work out and he changed direction mid way. The prompt equation did not add value to analyzing even statement 1 alone though we figure that out after analyzing it using the prompt equation. It's still equivalent to knowing that r is an integer.

Though it is absolutely fine if the prompt info is useful only for one statement. e.g. often you are given in the prompt that xy is not 0 implying that neither x nor y is 0.
x and/or y might be in the denominator in only one statement. It would be fine.

nitesh50 · Answer

If we divide the statement #1 equation by the prompt equation, we get  /  =r

While analysing this statement, I got this equation. Then can't we say that a+b is irrelevant to the reasoning.
if c and d are equal to 5 and 3 then answer is NO
if c and d are equal to 6 and 3 then answer is yes.

I feel the answer should be E.

Can any expert please comment?
 Bunuel 
 gmatbusters 
 chetan2u 
 Gladiator59

chetan2u · Answer

May be c=d, then r is 0, and so you cannot divide the equation the way you have done..
So do not divide or cross multiply when you are not sure if the variables are non zero or not..

(a+b)(c-d)=r.....(a+b)^2(c-d)^2=r^2=(a+b)(c+d).......(a+b)(c-d)^2=c+d ..
Now from statement II, a+b is square.... a+b=x^4*y^6*z^2=(x^2y^3z)^2 ..
Thus  c+d=(a+b)(c-d)^2=(x^2y^3z(c-d))^2 ..
Taking square root on both sides.. √(c+d)=x^2y^3z(c-d) 
As all variables are integer.. √(c+d) is an integer..

nitesh50 · Answer

Hi  chetan2u

So can I then imply that if the argument had stated that c does not equal to d, the answer would be E?.

Regards
Nitesh

gr8sars · Answer

Hey , 
There is 1 other method ( which is much quicker) to solve this question although I am not sure if it can be done or not .

Using statement 1.) we have (a+b)(c+d) = r^2 .
Now , if we were to take root on both sides of the eqn --> \sqrt{ } = r .

( can we take roots on both sides of an eqn comprising of variables without thinking of thinking of +ve or -ve sign ? )

Anyway , if the above step is correct , then using stmnt 2.) we have (a+b) = x^4 * y^6 * z^2 .

Combining both , we get --> r = \sqrt{ } ===> r = (x^2 * y^3 * z ) * \sqrt{ (c+d)}.
Since r is an integer , so are x.y and z . This implies that \sqrt{(c+d)} is also an integer .

Another query : Is the last implication sentence accurate ? Can \sqrt{(c+d)} not be a fractional/decimal value ?

SAHILBISHT0603 · Answer

From (1)
(a+b)=(c+d)=r
r*r=r^2

also as given (a+b)(c-d)=r ; so (c-d)=1

if c=8 & d=7 then c-d=1 but sqrt(c+d) not an integer
but
if c=5 & d=4 then c-d=1 but sqrt(c+d) is an integer

so (1) not sufficient

(2) says nothing about c-d ;so (2) not sufficient
combining
(a+b)=(c+d)=x^2*y^6*z^2

sqrt(c+d) is an integer; since power of prime are even

so upon combing definite yes ;Hence (c)

bumpbot · Answer

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a, b, c, and d are positive integers. If (a + b)(c d) = r, where

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a, b, c, and d are positive integers. If (a + b)(c d) = r, where

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