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

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a, b, c, and d are positive integers. If (a + b) (c – d) = r  [#permalink]

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New post 16 Jan 2013, 14:31
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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.

Thought I'd seen 'em all...then got stumped with this one! Give it a try.
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Re: a, b, c, and d are positive integers. If (a + b) (c – d) = r  [#permalink]

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New post 07 Jan 2014, 00:18
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sambam wrote:
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.

Thought I'd seen 'em all...then got stumped with this one! Give it a try.


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)
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Re: a, b, c, and d are positive integers. If (a + b) (c – d) = r  [#permalink]

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New post 16 Jan 2013, 14:47
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sambam wrote:
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.

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 [(c + d)]/[(c - d)] = 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) = [r^2]/[(a + b))], 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 :-)
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Re: a, b, c, and d are positive integers. If (a + b) (c – d) = r  [#permalink]

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New post 06 Jan 2014, 10:04
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sambam wrote:
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.

Thought I'd seen 'em all...then got stumped with this one! Give it a try.


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

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Re: a, b, c, and d are positive integers. If (a + b) (c – d) = r  [#permalink]

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New post 02 Apr 2014, 15:43
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!
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Re: a, b, c, and d are positive integers. If (a + b) (c – d) = r  [#permalink]

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New post 02 Apr 2014, 21:53
jlgdr wrote:
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


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.
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a, b, c, and d are positive integers. If (a + b) (c – d) = r  [#permalink]

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New post 28 Dec 2018, 22:43
sambam wrote:
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.

Thought I'd seen 'em all...then got stumped with this one! Give it a try.






If we divide the statement #1 equation by the prompt equation, we get [(c + d)]/[(c - d)] =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?
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Re: a, b, c, and d are positive integers. If (a + b) (c – d) = r  [#permalink]

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New post 28 Dec 2018, 23:05
nitesh50 wrote:
sambam wrote:
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.

Thought I'd seen 'em all...then got stumped with this one! Give it a try.






If we divide the statement #1 equation by the prompt equation, we get [(c + d)]/[(c - d)] =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
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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..
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Re: a, b, c, and d are positive integers. If (a + b) (c – d) = r  [#permalink]

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New post 28 Dec 2018, 23:16
chetan2u wrote:
nitesh50 wrote:
sambam wrote:
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.

Thought I'd seen 'em all...then got stumped with this one! Give it a try.






If we divide the statement #1 equation by the prompt equation, we get [(c + d)]/[(c - d)] =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
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chetan2u
Gladiator59


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..





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
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Re: a, b, c, and d are positive integers. If (a + b) (c – d) = r   [#permalink] 28 Dec 2018, 23:16
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