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1 and 8 are the first two natural numbers for which 1+2+3+...n is a pe

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honeyrai
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1 and 8 are the first two natural numbers for which 1+2+3+...n is a pe [#permalink] New post   19 Feb 2010, 03:56
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1 and 8 are the first two natural numbers for which 1+2+3+...n is a perfect square. Which number is 4th such number?



Sorry guys, I don't have any options in this question. Its a straight problem solving question.



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Re: 1 and 8 are the first two natural numbers for which 1+2+3+...n is a pe [#permalink] New post   19 Feb 2010, 06:37
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The sum of first \(n\) positive integers is given by the formula: \(\frac{n(n+1)}{2}\), we are told that this sum must be a perfect square, so we have:

\(\frac{n(n+1)}{2}=x^2\), obviously either \(n\) or \(n+1\) must be a perfect square itself, checking the perfect squares for \(n\) and \(n+1\) gives us following values for \(n\): 0, 1, 8, 49, 288... If you say that \(n=1\) is the first term, then the fourth term would be \(n=288\) --> \(\frac{n(n+1)}{2}=\frac{288(288+1)}{2}=144*289=12^2*17^2=x^2\).

Hope it helps.
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Re: 1 and 8 are the first two natural numbers for which 1+2+3+...n is a pe [#permalink] New post   19 Feb 2010, 06:59
Great reply.

But then the question is that how did you arrive at the values for n i.e. 0,1,8,49, 288... This is the main stumble block.

And why do you say that most probably either n or n+1 must be a perfect square.
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Re: 1 and 8 are the first two natural numbers for which 1+2+3+...n is a pe [#permalink] New post   19 Feb 2010, 07:17
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honeyrai wrote:
Great reply.

But then the question is that how did you arrive at the values for n i.e. 0,1,8,49, 288... This is the main stumble block.

And why do you say that most probably either n or n+1 must be a perfect square.


\(x=\sqrt{\frac{n(n+1)}{2}}\), as \(x\) is an integer, either \(n\) or \(n+1\) MUST be a perfect square (not probably). If \(n\) is a perfect square, then \(\frac{n+1}{2}\) is also a perfect square and vise-versa. After this I did by plugging numbers, which is in this case the fastest way.
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Re: 1 and 8 are the first two natural numbers for which 1+2+3+...n is a pe [#permalink] New post   19 Feb 2010, 07:46
+1 for you.

This question is very time consuming & do you think that these kind of questions can be asked on GMAT.

My last quant score was 48. I want to increase it to 51. So I need some really tough problem solving. Can you suggest me some material for that?
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Re: 1 and 8 are the first two natural numbers for which 1+2+3+...n is a pe [#permalink] New post   19 Feb 2010, 11:50
honeyrai wrote:
+1 for you.

This question is very time consuming & do you think that these kind of questions can be asked on GMAT.

My last quant score was 48. I want to increase it to 51. So I need some really tough problem solving. Can you suggest me some material for that?


48 is nevertheless a good score. Most GMAT questions aim at focusing on your ability to derive an appropriate equation or relationship between the given conditions. They are appropriately solvable within constraints of the math fundamentals. They don't expect you to derive a formula to calculate factorials and the like. So don't worry, am very sure, this kind of a question will never appear on the GMAT. Even if plugging in numbers are involved, am very sure, for values of n such as 1 & 8 given in this question in particular, the next values will be much closer and not thrown far apart like 49 and 288.
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Re: 1 and 8 are the first two natural numbers for which 1+2+3+...n is a pe [#permalink] New post   19 Feb 2010, 14:00
Hey Barney! Thanks for your soothing words!
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Re: 1 and 8 are the first two natural numbers for which 1+2+3+...n is a pe [#permalink] New post   19 Feb 2010, 15:29
Bunuel wrote:
honeyrai wrote:
Great reply.

But then the question is that how did you arrive at the values for n i.e. 0,1,8,49, 288... This is the main stumble block.

And why do you say that most probably either n or n+1 must be a perfect square.


\(x=\sqrt{\frac{n(n+1)}{2}}\), as \(x\) is an integer, either \(n\) or \(n+1\) MUST be a perfect square (not probably). If \(n\) is a perfect square, then \(\frac{n+1}{2}\) is also a perfect square and vise-versa. After this I did by plugging numbers, which is in this case the fastest way.


say n =64 which is a perfect square n+1/2 is 65/2 which is not even an integer....i am confused
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Re: 1 and 8 are the first two natural numbers for which 1+2+3+...n is a pe [#permalink] New post   19 Feb 2010, 22:14
ISBtarget wrote:
say n =64 which is a perfect square n+1/2 is 65/2 which is not even an integer....i am confused

Thats the key. Look for numbers which are perfect square & which are odd. Don't look for even perfect squares. This will help you in quickly hopping numbers & reaching the number fast.

8 matches with perfect square 9 because 8/2 & 9 both are perfect squares & 8&9 are also the consecutive numbers. So start looking for odd perfect squares after 9. Check 25, 49(yes), 81, 121, 169, 225, 289 (yes)...

This is how you can really solve fast.



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Re: 1 and 8 are the first two natural numbers for which 1+2+3+...n is a pe [#permalink]
19 Feb 2010, 22:14
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