GMAT Changed on April 16th - Read about the latest changes here

 It is currently 20 May 2018, 20:30

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
Director
Joined: 01 Jan 2008
Posts: 602
X is called a triangular number if X = 1 + 2 + ... + n for [#permalink]

### Show Tags

01 Oct 2008, 06:15
1
KUDOS
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?

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Manager
Joined: 22 Sep 2008
Posts: 117
Re: walker type of problem [#permalink]

### Show Tags

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.
Intern
Joined: 29 Sep 2008
Posts: 46
Re: walker type of problem [#permalink]

### Show Tags

01 Oct 2008, 07:59
1
KUDOS
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
Director
Joined: 01 Jan 2008
Posts: 602
Re: walker type of problem [#permalink]

### Show Tags

01 Oct 2008, 08:14
Nice solution, aim2010. I solved it exactly the same way.
Manager
Joined: 30 Sep 2008
Posts: 111
Re: walker type of problem [#permalink]

### Show Tags

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
Joined: 29 Sep 2008
Posts: 46
Re: walker type of problem [#permalink]

### Show Tags

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.
CEO
Joined: 17 Nov 2007
Posts: 3486
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Re: walker type of problem [#permalink]

### Show Tags

01 Oct 2008, 14:10
Thanks maraticus for a good question and aim2010 for a nice explanation
+1+1
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Manager
Joined: 30 Sep 2008
Posts: 111
Re: walker type of problem [#permalink]

### Show Tags

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
VP
Joined: 18 May 2008
Posts: 1182
Re: walker type of problem [#permalink]

### Show Tags

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

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Re: walker type of problem   [#permalink] 01 Oct 2008, 17:38
Display posts from previous: Sort by

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

Moderator: chetan2u

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.