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by rajivmatta » Sat Dec 29, 2007 2:14 pm
Q] If x,y,a,b are all positive Integers and x^3 + y^3 = a^3 + b^3, then is xy>89

1] a^3 + b^3 = 1729
2] x^2 = x^3


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by StarDust845 » Sat Dec 29, 2007 9:48 pm
Is the answer A?

Calista.
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by rajivmatta » Sun Dec 30, 2007 9:34 am
StarDust845 wrote:Is the answer A?

Calista.
First condition is not sufficient

Here is a clue
a^3 + b^3 = x^3+y^3 = 1729...a,b could be 12,1 or 10,9
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by samirpandeyit62 » Sun Dec 30, 2007 11:24 pm
Q] If x,y,a,b are all positive Integers and x^3 + y^3 = a^3 + b^3, then is xy>89

since all the nos are +ve so we have

1] a^3 + b^3 = 1729
largest +ve nos that can be considered here is 12 coz 12 ^3 =1728(we need sum as 1729)
hence we have 12,1 are possible values
we can also evaluate some other cubes like 10, 9 which will also give the same sum

8^3 & 7^3 and cubes of all numbers below 9 will not work coz we cannot pair them with any other value within 12

so we have two possibilities 12,1 (12) & 10,9 (90) INSUFF


2] x^2 = x^3

only nos which satisfies this is 1

hoever value of y can be anything INSUFF

combine x^3+y^3 =1729 & x =1 so y must be 12 here to satisfy stmt 1

SUFF

C
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Samir
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by rajivmatta » Mon Dec 31, 2007 3:01 pm
You cracked it Samir

OA C...
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