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# X is called a triangular number if X = 1 + 2 + ... + n for

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X is called a triangular number if X = 1 + 2 + ... + n for [#permalink]

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01 Oct 2008, 06:15
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X is called a triangular number if X = 1 + 2 + ... + n for some positive integer n
X is called a perfect square if X = k^2 for some integer k

The first X which is both a triangular number and a perfect square is 1, the second 36. What's the third one?
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Re: walker type of problem [#permalink]

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01 Oct 2008, 06:46
i think we should try this type of question by OPTION only..

Ultimately in gmat we are going to have option , just apply 4 option one by one , we will get answer definately

* i think we need to find number which is cube of some number and following number is square of some number.
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Re: walker type of problem [#permalink]

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01 Oct 2008, 07:59
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a number that satisfies both requirements can be expressed as n(n+1)/2 and k^2.

Since it is the same number:

n(n+1)/2=k^2
=> n(n+1) = 2 * k^2

careful observation reveals that we are looking for a positive integer that can be expressed as a product of two consecutive positive integers and also as double of a perfect square.

quick things to realize:
1. two consecutive positive integers would never have a common factor (except 1 which we can ignore)
2. two consecutive positive integers would have one odd and one even integer
3. odd integer has to be a perfect square
4. 2 is a factor of only the even integer which would cancel out the 2 from the other side of the equation
5. after the factor 2 is taken out of the even integer, it is also a perfect square

now lets start looking for odd perfect squares whose adjacent even integer is double of a perfect square.

1 and 2 - our number is 1 * 2 /2 = 1
9 and 8 - our number is 9 * 8 /2 = 36
25 - reject, 24 or 26 do not satisfy
49 and 50 - our number is 49 * 50 /2 = 1225
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Re: walker type of problem [#permalink]

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01 Oct 2008, 08:14
Nice solution, aim2010. I solved it exactly the same way.
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Re: walker type of problem [#permalink]

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01 Oct 2008, 08:22
aim2010 wrote:
a number that satisfies both requirements can be expressed as n(n+1)/2 and k^2.

Since it is the same number:

n(n+1)/2=k^2
=> n(n+1) = 2 * k^2

careful observation reveals that we are looking for a positive integer that can be expressed as a product of two consecutive positive integers and also as double of a perfect square.

quick things to realize:
1. two consecutive positive integers would never have a common factor (except 1 which we can ignore)
2. two consecutive positive integers would have one odd and one even integer
3. odd integer has to be a perfect square
4. 2 is a factor of only the even integer which would cancel out the 2 from the other side of the equation
5. after the factor 2 is taken out of the even integer, it is also a perfect square

now lets start looking for odd perfect squares whose adjacent even integer is double of a perfect square.

1 and 2 - our number is 1 * 2 /2 = 1
9 and 8 - our number is 9 * 8 /2 = 36
25 - reject, 24 or 26 do not satisfy
49 and 50 - our number is 49 * 50 /2 = 1225

for these similar questions, I won't go so far, esp to many calculations, up to 49(!!!), should we just replace the number from the choices and

check n(n+1) = 2 * k^2 if n is integer
Intern
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Re: walker type of problem [#permalink]

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01 Oct 2008, 08:50
lylya4 wrote:

for these similar questions, I won't go so far, esp to many calculations, up to 49(!!!), should we just replace the number from the choices and

check n(n+1) = 2 * k^2 if n is integer

it makes sense to a certain extent, though there are two reasons I would not test answer choices:
1. Imagine checking 5 four digit numbers if they are perfect squares and if all of them are, then doubling them and checking their factors to come up with two consecutive integers as two exhaustive factors
2. checking answer choices at times gives me a feeling that maybe i missed something, what if i do a mistake and two choices seem to satisfy.

moreover, for going till 49 we just went till 7 after counting 1, 3 and 5. remember we were looking for odd perfect squares. IMO, its up to each person's choice. bottom line is to not bang your head on something for 4 minutes unless you have a gut feeling that you are going in the right direction.
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Re: walker type of problem [#permalink]

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01 Oct 2008, 14:10
Thanks maraticus for a good question and aim2010 for a nice explanation
+1+1
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Manager
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Re: walker type of problem [#permalink]

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01 Oct 2008, 16:25
aim2010 wrote:
lylya4 wrote:

for these similar questions, I won't go so far, esp to many calculations, up to 49(!!!), should we just replace the number from the choices and

check n(n+1) = 2 * k^2 if n is integer

it makes sense to a certain extent, though there are two reasons I would not test answer choices:
1. Imagine checking 5 four digit numbers if they are perfect squares and if all of them are, then doubling them and checking their factors to come up with two consecutive integers as two exhaustive factors
2. checking answer choices at times gives me a feeling that maybe i missed something, what if i do a mistake and two choices seem to satisfy.

moreover, for going till 49 we just went till 7 after counting 1, 3 and 5. remember we were looking for odd perfect squares. IMO, its up to each person's choice. bottom line is to not bang your head on something for 4 minutes unless you have a gut feeling that you are going in the right direction.

i still think its easier to check the choices

remember k^2 = X

So n(n+1) = 2X <=> n^2 + 2n - 2X = 0, you replace X with the answer choice and solve the equation, if n is integer, take X

much faster
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Re: walker type of problem [#permalink]

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01 Oct 2008, 17:38
That seems to be a better approach
lylya4 wrote:
aim2010 wrote:
lylya4 wrote:

for these similar questions, I won't go so far, esp to many calculations, up to 49(!!!), should we just replace the number from the choices and

check n(n+1) = 2 * k^2 if n is integer

it makes sense to a certain extent, though there are two reasons I would not test answer choices:
1. Imagine checking 5 four digit numbers if they are perfect squares and if all of them are, then doubling them and checking their factors to come up with two consecutive integers as two exhaustive factors
2. checking answer choices at times gives me a feeling that maybe i missed something, what if i do a mistake and two choices seem to satisfy.

moreover, for going till 49 we just went till 7 after counting 1, 3 and 5. remember we were looking for odd perfect squares. IMO, its up to each person's choice. bottom line is to not bang your head on something for 4 minutes unless you have a gut feeling that you are going in the right direction.

i still think its easier to check the choices

remember k^2 = X

So n(n+1) = 2X <=> n^2 + 2n - 2X = 0, you replace X with the answer choice and solve the equation, if n is integer, take X

much faster
Re: walker type of problem   [#permalink] 01 Oct 2008, 17:38
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