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3

$begingroup$

I have tried to solve few number series problems but this one is little difficult for me to solve. kindly help me to solve this.

-1,35,143,323,?

Thanks in advance.

number-sequence

share|improve this question

asked Mar 21 at 13:45

T. ArunkumarT. Arunkumar

213

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put on hold as off-topic by gabbo1092, w l, JonMark Perry, Glorfindel, El-Guest 9 hours ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This looks like a puzzle you found elsewhere. For content you did not create yourself, proper attribution is required. If you have permission to repost this, please edit to include (at minimum) where it came from, then vote to reopen. Posts which use someone else's content without attribution are generally deleted." – gabbo1092, w l, JonMark Perry, Glorfindel, El-Guest

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hm.. Maybe the first number should be $6$ instead of $-1$?
  $endgroup$
  – athin
  Mar 21 at 13:47
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  This looks like this puzzle may have been taken from somewhere (v.s. one you created) - if so please provide attribution to the source. And welcome to Puzzling SE!
  $endgroup$
  – PartyHatPanda
  Mar 21 at 14:05
  

  
  
  
  

add a comment | 

3

$begingroup$

I have tried to solve few number series problems but this one is little difficult for me to solve. kindly help me to solve this.

-1,35,143,323,?

Thanks in advance.

number-sequence

share|improve this question

asked Mar 21 at 13:45

T. ArunkumarT. Arunkumar

213

New contributor

T. Arunkumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

put on hold as off-topic by gabbo1092, w l, JonMark Perry, Glorfindel, El-Guest 9 hours ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This looks like a puzzle you found elsewhere. For content you did not create yourself, proper attribution is required. If you have permission to repost this, please edit to include (at minimum) where it came from, then vote to reopen. Posts which use someone else's content without attribution are generally deleted." – gabbo1092, w l, JonMark Perry, Glorfindel, El-Guest

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hm.. Maybe the first number should be $6$ instead of $-1$?
  $endgroup$
  – athin
  Mar 21 at 13:47
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  This looks like this puzzle may have been taken from somewhere (v.s. one you created) - if so please provide attribution to the source. And welcome to Puzzling SE!
  $endgroup$
  – PartyHatPanda
  Mar 21 at 14:05
  

  
  
  
  

add a comment | 

$begingroup$

I have tried to solve few number series problems but this one is little difficult for me to solve. kindly help me to solve this.

-1,35,143,323,?

Thanks in advance.

number-sequence

share|improve this question

asked Mar 21 at 13:45

T. ArunkumarT. Arunkumar

213

New contributor

T. Arunkumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

-1,35,143,323,?

share|improve this question

asked Mar 21 at 13:45

T. ArunkumarT. Arunkumar

213

New contributor

T. Arunkumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

asked Mar 21 at 13:45

T. ArunkumarT. Arunkumar

213

New contributor

T. Arunkumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 21 at 13:45

T. ArunkumarT. Arunkumar

213

New contributor

T. Arunkumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 21 at 13:45

T. ArunkumarT. Arunkumar

213

2 Answers
2

active

oldest

votes

10

$begingroup$

I think I got this:

0*0 -1 = -1

6*6 -1 = 35

12*12-1 = 143

18*18 -1 = 323

24*24 -1 = 575

or:

(6*0)^2 -1 = -1

(6*1)^2 -1 = 35

(6*2)^2-1 = 143

(6*3)^2 -1 = 323

(6*4)^2 -1 = 575

share|improve this answer

edited Mar 21 at 14:25

answered Mar 21 at 13:50

BediBedi

11615

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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Or Maybe there is a Shorter way that im missing
  $endgroup$
  – Bedi
  Mar 21 at 13:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My comment might be confusing after the edit so I removed it, but +1 from me now
  $endgroup$
  – PunPun1000
  Mar 21 at 14:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Or (-1)x1, 5x7, 11x13, 17x19, 23x25, ...
  $endgroup$
  – OHO
  Mar 21 at 16:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @OHO That's neat. It holds true in math that $x^2 = (x + y)(x - y) + y^2$. In this instance $y$ is just 1 in every case; so it's the same equation, just a different representation. You may have known that already, just thought it was interesting.
  $endgroup$
  – Lord Farquaad
  Mar 21 at 20:53
  

  
  
  
  

add a comment | 

5

$begingroup$

Bedi’s got the idea. Another way to write it would be:

(6n)^2 - 1 where n goes from 0 to 4

So, the next number is

(6*4)^2 - 1 = 575

share|improve this answer

answered Mar 21 at 14:10

arbitrahjarbitrahj

995113

$endgroup$

add a comment | 

10

$begingroup$

I think I got this:

0*0 -1 = -1

6*6 -1 = 35

12*12-1 = 143

18*18 -1 = 323

24*24 -1 = 575

or:

(6*0)^2 -1 = -1

(6*1)^2 -1 = 35

(6*2)^2-1 = 143

(6*3)^2 -1 = 323

(6*4)^2 -1 = 575

share|improve this answer

edited Mar 21 at 14:25

answered Mar 21 at 13:50

BediBedi

11615

New contributor

Bedi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Or Maybe there is a Shorter way that im missing
  $endgroup$
  – Bedi
  Mar 21 at 13:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My comment might be confusing after the edit so I removed it, but +1 from me now
  $endgroup$
  – PunPun1000
  Mar 21 at 14:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Or (-1)x1, 5x7, 11x13, 17x19, 23x25, ...
  $endgroup$
  – OHO
  Mar 21 at 16:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @OHO That's neat. It holds true in math that $x^2 = (x + y)(x - y) + y^2$. In this instance $y$ is just 1 in every case; so it's the same equation, just a different representation. You may have known that already, just thought it was interesting.
  $endgroup$
  – Lord Farquaad
  Mar 21 at 20:53
  

  
  
  
  

add a comment | 

10

$begingroup$

I think I got this:

0*0 -1 = -1

6*6 -1 = 35

12*12-1 = 143

18*18 -1 = 323

24*24 -1 = 575

or:

(6*0)^2 -1 = -1

(6*1)^2 -1 = 35

(6*2)^2-1 = 143

(6*3)^2 -1 = 323

(6*4)^2 -1 = 575

share|improve this answer

edited Mar 21 at 14:25

answered Mar 21 at 13:50

BediBedi

11615

New contributor

Bedi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Or Maybe there is a Shorter way that im missing
  $endgroup$
  – Bedi
  Mar 21 at 13:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My comment might be confusing after the edit so I removed it, but +1 from me now
  $endgroup$
  – PunPun1000
  Mar 21 at 14:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Or (-1)x1, 5x7, 11x13, 17x19, 23x25, ...
  $endgroup$
  – OHO
  Mar 21 at 16:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @OHO That's neat. It holds true in math that $x^2 = (x + y)(x - y) + y^2$. In this instance $y$ is just 1 in every case; so it's the same equation, just a different representation. You may have known that already, just thought it was interesting.
  $endgroup$
  – Lord Farquaad
  Mar 21 at 20:53
  

  
  
  
  

add a comment | 

$begingroup$

I think I got this:

0*0 -1 = -1

6*6 -1 = 35

12*12-1 = 143

18*18 -1 = 323

24*24 -1 = 575

or:

(6*0)^2 -1 = -1

(6*1)^2 -1 = 35

(6*2)^2-1 = 143

(6*3)^2 -1 = 323

(6*4)^2 -1 = 575

share|improve this answer

edited Mar 21 at 14:25

answered Mar 21 at 13:50

BediBedi

11615

New contributor

Bedi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this answer

edited Mar 21 at 14:25

answered Mar 21 at 13:50

BediBedi

11615

New contributor

Bedi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Mar 21 at 13:50

BediBedi

11615

New contributor

Bedi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Mar 21 at 13:50

BediBedi

11615

5

$begingroup$

Bedi’s got the idea. Another way to write it would be:

(6n)^2 - 1 where n goes from 0 to 4

So, the next number is

(6*4)^2 - 1 = 575

share|improve this answer

answered Mar 21 at 14:10

arbitrahjarbitrahj

995113

$endgroup$

add a comment | 

5

$begingroup$

Bedi’s got the idea. Another way to write it would be:

(6n)^2 - 1 where n goes from 0 to 4

So, the next number is

(6*4)^2 - 1 = 575

share|improve this answer

answered Mar 21 at 14:10

arbitrahjarbitrahj

995113

$endgroup$

add a comment | 

$begingroup$

Bedi’s got the idea. Another way to write it would be:

(6n)^2 - 1 where n goes from 0 to 4

So, the next number is

(6*4)^2 - 1 = 575

share|improve this answer

answered Mar 21 at 14:10

arbitrahjarbitrahj

995113

$endgroup$

share|improve this answer

answered Mar 21 at 14:10

arbitrahjarbitrahj

995113

answered Mar 21 at 14:10

arbitrahjarbitrahj

995113

answered Mar 21 at 14:10

arbitrahjarbitrahj

995113