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Re: a, b, c, and d are positive integers. If (a + b)(c d) = r, where [#permalink]
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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, where [#permalink]
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, where [#permalink]
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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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Re: a, b, c, and d are positive integers. If (a + b)(c d) = r, where [#permalink]
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, where [#permalink]
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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, where [#permalink]
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, where [#permalink]
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{[ (a+b)(c+d) ]} = 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{[ (x^4 * y^6 * z^2 )(c+d) ]} ===> 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 ?
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Re: a, b, c, and d are positive integers. If (a + b)(c d) = r, where [#permalink]
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)
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Re: a, b, c, and d are positive integers. If (a + b)(c d) = r, where [#permalink]
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