DS-Tough one?

Question

I don't know whether it is forbidden to post questions from other sites. However, this is from https://www.manhattangmat.com . I apologize in advance if I violate any rule. Here comes the question:

FN · Answer

C it is....

(a+b)(c-d)=r ; an integer


i) (a+b)(c+d)=r^2

doesnt say much about sqrt(c+d)..insuff

II) (a+b)= perfect square! prime^even is always perfect square!

not much about sqrt(c+d)...Insuff

combining them together...

I) (a+b)=r/(c-d) <---from problem stem

r/(c-d) * (c+d)=r^2; (c+d)/(c-d)=r an integer!

now lets plug back in.....

problem stem says (a+b)(c-d)=(c+d)/(c-d)
muliply both sides with (c-d), we have (a+b)(c-d)^2= (c+d)

we know that (a+b) is a perfect square from II), (c+d)^2 is also integer, 

so now take Sqrt(perfect.sq * (c-d)^2)=sqrt(c+d)

we know that sqrt(c+d) is an integer!

sparky · Answer

yeah, nice solution

it should be C.

HIMALAYA · Answer

i also think C.
lets do short cut since we know i and ii independently doesnot answer the question. so the possible answers are eith C or E.

from i and ii,
(a+b)(c+d)=r^2
(a+b)(c+d) = (a+b)(c-d)(a+b)(c-d)
(c+d)=(a+b)(c-d)^2
sqrt(c+d) = sqrt 
sqrt(c+d) = (x^2 y^3 z) (c-d). this must be an integer.

HowManyToGo · Answer

C.

From 1 ) c + d = (r ^ 2)/(a + b)
 = (a+b)^2 * (c-d)^2 / ( a+ b)
 = (a+b)*(c-d)^2
 (a + b) can or cannot be a perfect square ,hence insufficient.

From 2) (a + b) is a perfect square. ( still no clue about c + d )

Using 1 and 2. c + d = (a+b) (c -d ) ^2
 = (x^2*y^3*z)^2 * (c - d ) ^2
 = ((x^2*y^3*z) * (c - d )) ^2.( a perfect square)

Hence sufficient.

HMTG.

DS-Tough one?

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