DS-Tough one? : DS Archive
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# DS-Tough one?

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Senior Manager
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06 Jun 2005, 13:07
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This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I don't know whether it is forbidden to post questions from other sites. However, this is from http://www.manhattangmat.com . I apologize in advance if I violate any rule.
Here comes the question:
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06 Jun 2005, 16:19
C it is....

(a+b)(c-d)=r ; an integer

i) (a+b)(c+d)=r^2

II) (a+b)= perfect square! prime^even is always perfect square!

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!
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06 Jun 2005, 17:12
yeah, nice solution

it should be C.
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06 Jun 2005, 17:16
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[(x^4 y^6 z^2) (c-d)^2]
sqrt(c+d) = (x^2 y^3 z) (c-d). this must be an integer.
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07 Jun 2005, 07:58
thearch wrote:
I don't know whether it is forbidden to post questions from other sites. However, this is from http://www.manhattangmat.com . I apologize in advance if I violate any rule.
Here comes the question:

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.
Re: DS-Tough one?   [#permalink] 07 Jun 2005, 07:58
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