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# There are three consecutive #s such that the square of the

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Senior Manager
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There are three consecutive #s such that the square of the [#permalink]  03 Mar 2004, 16:31
There are three consecutive #s such that the square of the middle # exceeds the difference of the squares of the first and last #s by 60. What is the last of the three #s?

10
14
11
17

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Pls include reasoning along with all answer posts.
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Director
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Ans: 11

Let the numbers be a, a+1 and a+2

(a+1)^2 = 60 + [(a+2)^2 - a^2]

Solve for "a"
Manager
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Try this.. [#permalink]  03 Mar 2004, 17:03

I think if you tried a-1, a and a+1 your calulations will be much easier..
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Re: Try this.. [#permalink]  03 Mar 2004, 18:48
mantha wrote:

I think if you tried a-1, a and a+1 your calulations will be much easier..

Mantha:

Thank you for the tip.
This is the power of collaborative learning!

Last edited by kpadma on 04 Mar 2004, 04:39, edited 1 time in total.
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Kpadma, please note that the question is about the DIFFERENCE BETWEEN THE FIRST AND THE LAST NUMBER not the other way round.If you try the values that you have suggested you 'll see that 9,10,11 9^2-11^2=-40 so the square of the middle number exceeds the difference by 140, not 60. IMO, the numbers are -11,-10,-9 so the answer should be E)
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why is mantha's method (yielding 10 and -6 as solutions and thereby a+1 becomes 11 as answer ) not correct ? I didn't understand.
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BG wrote:
Kpadma, please note that the question is about the DIFFERENCE BETWEEN THE FIRST AND THE LAST NUMBER not the other way round.

Yes, you've a point. I've also noticed that issue, when I solved the problem.

But, Is it not true that the difference between two number is always postive
or zero. For exampble, "difference between A and B" is | A - B |, whereas
"A exceeds B" is A - B.

I could be wrong, so it is worth clearing up this confusion.

Dear Praetorian or Akamai,
Could you help us with this issue?
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