Find the next value of this number series [on hold]Number Series - 1,20,171Find the next number in this seriesComplete the Number Sequence SeriesWhat's the next number in this sequence?Fill in the blanks of this number sequenceWhat is the number in the square with “?”?What is the next number is this sequence?Can you find me using a number sequence?Sequence/Series - Find $a_n$ with only $n$What is the next number in the given series?
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Find the next value of this number series [on hold]
Number Series - 1,20,171Find the next number in this seriesComplete the Number Sequence SeriesWhat's the next number in this sequence?Fill in the blanks of this number sequenceWhat is the number in the square with “?”?What is the next number is this sequence?Can you find me using a number sequence?Sequence/Series - Find $a_n$ with only $n$What is the next number in the given series?
$begingroup$
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
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T. Arunkumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
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
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
$begingroup$
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
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.
$endgroup$
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
number-sequence
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.
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.
New contributor
T. Arunkumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Mar 21 at 13:45
![](https://i.stack.imgur.com/tO3Mi.jpg?s=32&g=1)
![](https://i.stack.imgur.com/tO3Mi.jpg?s=32&g=1)
T. ArunkumarT. Arunkumar
213
213
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T. Arunkumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
T. Arunkumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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T. Arunkumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
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
$begingroup$
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 |
$begingroup$
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
$begingroup$
Hm.. Maybe the first number should be $6$ instead of $-1$?
$endgroup$
– athin
Mar 21 at 13:47
$begingroup$
Hm.. Maybe the first number should be $6$ instead of $-1$?
$endgroup$
– athin
Mar 21 at 13:47
4
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
$begingroup$
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 |
2 Answers
2
active
oldest
votes
$begingroup$
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
New contributor
Bedi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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 |
$begingroup$
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
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
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
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
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.
$endgroup$
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
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.
edited Mar 21 at 14:25
New contributor
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answered Mar 21 at 13:50
![](https://lh4.googleusercontent.com/-GQKxhs8o_jw/AAAAAAAAAAI/AAAAAAAAAAA/APUIFaPbZHN20YDS-gxDPXnIdPDOog7Esg/mo/photo.jpg?sz=32)
![](https://lh4.googleusercontent.com/-GQKxhs8o_jw/AAAAAAAAAAI/AAAAAAAAAAA/APUIFaPbZHN20YDS-gxDPXnIdPDOog7Esg/mo/photo.jpg?sz=32)
BediBedi
11615
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.
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.
Bedi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
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 |
$begingroup$
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
$begingroup$
Or Maybe there is a Shorter way that im missing
$endgroup$
– Bedi
Mar 21 at 13:51
$begingroup$
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
$begingroup$
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
1
$begingroup$
Or (-1)x1, 5x7, 11x13, 17x19, 23x25, ...
$endgroup$
– OHO
Mar 21 at 16:44
$begingroup$
Or (-1)x1, 5x7, 11x13, 17x19, 23x25, ...
$endgroup$
– OHO
Mar 21 at 16:44
1
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
$begingroup$
@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 |
$begingroup$
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
$endgroup$
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
$endgroup$
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
$endgroup$
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
answered Mar 21 at 14:10
![](https://i.stack.imgur.com/DaTeM.png?s=32&g=1)
![](https://i.stack.imgur.com/DaTeM.png?s=32&g=1)
arbitrahjarbitrahj
995113
995113
add a comment |
add a comment |
$begingroup$
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