Natural numbers

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by Brent@GMATPrepNow » Tue Aug 27, 2013 7:17 am
gmatkkvinu wrote:1 and 8 are the first two natural number for which 1+2+3+.........+n is a perfect square.which number is the fourth such number?
I don't believe that this question is of GMAT quality.
Unless I'm missing something, it requires EITHER too much brute force (adding lots of numbers while looking for squares) OR advanced math concepts that are not tested on the GMAT.

Also, if this were a GMAT question, the answer choices would be provided, in which case we could test the different options.

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by Brent@GMATPrepNow » Tue Aug 27, 2013 7:55 am
gmatkkvinu wrote:1 and 8 are the first two natural number for which 1+2+3+.........+n is a perfect square.which number is the fourth such number?
I still maintain that this is out of scope for the GMAT, but here's my solution nevertheless. . .

We might start with this formula: 1+2+3...+n = (n)(n+1)/2

So, we're looking for values of n such that (n)(n+1)/2 is a perfect square (i.e., the square of an integer).
As the question shows, when n = 1, we get (n)(n+1)/2 = 1, which is a perfect square.
Similarly, when n = 8, we get (n)(n+1)/2 = 36, which is a perfect square.

IMPORTANT: First notice that n and n+1 are consecutive integers. Also notice that something nice happens when n = 8. When we divide n (8) by 2, we get 4 (which is a perfect square) and n+1 (9) is already a perfect square.

So, at this point, we need only check consecutive integers where one of the integers is already a square and the other integer, when divided by 2, becomes a square.

Check 15 and 16: 16 is a square, but 15/2 is not a square. Keep checking.
Check 16 and 17: 16 is a square, but 17/2 is not a square. Keep checking.
Check 24 and 25: 25 is a square, but 24/2 is not a square. Keep checking.
Check 25 and 26: 25 is a square, but 26/2 is not a square. Keep checking.

IMPORTANT: From now on, I won't check squares that are even, because the number that's 1 greater will be ODD, and odd/2 cannot be a perfect square. Likewise, the number that's 1 less will be ODD, and odd/2 cannot be a perfect square.

Check 48 and 49: 49 is a square, but 48/2 is not a square. Keep checking.
Check 49 and 50: 49 is a square, AND 50/2 is a square. So, 49 is the 3rd such number.
Check 80 and 81: 81 is a square, but 80/2 is not a square. Keep checking.
Check 81 and 82: 81 is a square, but 82/2 is not a square. Keep checking.
Check 120 and 121: 121 is a square, but 120/2 is not a square. Keep checking.
Check 121 and 122: 121 is a square, but 122/2 is not a square. Keep checking.
Check 168 and 169: 169 is a square, but 168/2 is not a square. Keep checking.
Check 169 and 170: 169 is a square, but 170/2 is not a square. Keep checking.
Check 224 and 225: 225 is a square, but 224/2 is not a square. Keep checking.
Check 225 and 226: 225 is a square, but 226/2 is not a square. Keep checking.
Check 288 and 289: 289 is a square, AND 288/2 is a square. So, n = 288 is the 4th such number.

There may be a faster (simpler) way, but that's my solution.

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Brent
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by Brent@GMATPrepNow » Tue Aug 27, 2013 7:56 am
gmatkkvinu wrote:1 and 8 are the first two natural number for which 1+2+3+.........+n is a perfect square.which number is the fourth such number?
Looks like my solution was posted twice somehow. I'm deleting one of them.

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by sanjoy18 » Fri Aug 30, 2013 8:56 am
1+2+3+..+n=n(n+1)/2= a square number
lets say n(n+1)/2= K^2
=> n(n+1)= 2 K^2
In LHS there is two consecutive natural number n and n+1. Therefore they don't have nay common factor.
since RHS is 2*square number.Therefore between n and n+1 one must be twice of a perfect square and other must be also a square number
since twice of a square number is even.another number must be odd.
so our task is to find one square number that is ODD and another 2* square number which must the neighbor of odd square number

3rd number.= 49 *50= 7^2*(2*5^2) >>> 3rd number 49
4th number= 289*288=17^2*(2*12^2)..hence 4th number is 288