Find the next value of this number seriesNumber 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
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
asked yesterday
T. ArunkumarT. Arunkumar
233
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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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$endgroup$
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
asked yesterday
T. ArunkumarT. Arunkumar
233
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.
$endgroup$
$begingroup$
Hm.. Maybe the first number should be $6$ instead of $-1$?
$endgroup$
– athin
yesterday
2
$begingroup$
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
yesterday
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
asked yesterday
T. ArunkumarT. Arunkumar
233
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
asked yesterday
T. ArunkumarT. Arunkumar
233
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 yesterday
T. ArunkumarT. Arunkumar
233
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 yesterday
T. ArunkumarT. Arunkumar
233
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 yesterday
T. ArunkumarT. Arunkumar
233
asked yesterday
T. ArunkumarT. Arunkumar
233
233
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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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$begingroup$
Hm.. Maybe the first number should be $6$ instead of $-1$?
$endgroup$
– athin
yesterday
2
$begingroup$
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
yesterday
add a comment |
$begingroup$
Hm.. Maybe the first number should be $6$ instead of $-1$?
$endgroup$
– athin
yesterday
2
$begingroup$
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
yesterday
$begingroup$
Hm.. Maybe the first number should be $6$ instead of $-1$?
$endgroup$
– athin
yesterday
$begingroup$
Hm.. Maybe the first number should be $6$ instead of $-1$?
$endgroup$
– athin
yesterday
2
2
$begingroup$
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
yesterday
$begingroup$
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
yesterday
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
answered yesterday
BediBedi
963
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
yesterday
$begingroup$
My comment might be confusing after the edit so I removed it, but +1 from me now
$endgroup$
– PunPun1000
yesterday
1
$begingroup$
Or (-1)x1, 5x7, 11x13, 17x19, 23x25, ...
$endgroup$
– OHO
yesterday
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
yesterday
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
answered yesterday
arbitrahjarbitrahj
630111
$endgroup$
add a comment |
Your Answer
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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
answered yesterday
BediBedi
963
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
yesterday
$begingroup$
My comment might be confusing after the edit so I removed it, but +1 from me now
$endgroup$
– PunPun1000
yesterday
1
$begingroup$
Or (-1)x1, 5x7, 11x13, 17x19, 23x25, ...
$endgroup$
– OHO
yesterday
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
yesterday
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
answered yesterday
BediBedi
963
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
yesterday
$begingroup$
My comment might be confusing after the edit so I removed it, but +1 from me now
$endgroup$
– PunPun1000
yesterday
1
$begingroup$
Or (-1)x1, 5x7, 11x13, 17x19, 23x25, ...
$endgroup$
– OHO
yesterday
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
yesterday
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
answered yesterday
BediBedi
963
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
answered yesterday
BediBedi
963
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 yesterday
answered yesterday
BediBedi
963
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 yesterday
BediBedi
963
answered yesterday
BediBedi
963
963
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
yesterday
$begingroup$
My comment might be confusing after the edit so I removed it, but +1 from me now
$endgroup$
– PunPun1000
yesterday
1
$begingroup$
Or (-1)x1, 5x7, 11x13, 17x19, 23x25, ...
$endgroup$
– OHO
yesterday
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
yesterday
add a comment |
$begingroup$
Or Maybe there is a Shorter way that im missing
$endgroup$
– Bedi
yesterday
$begingroup$
My comment might be confusing after the edit so I removed it, but +1 from me now
$endgroup$
– PunPun1000
yesterday
1
$begingroup$
Or (-1)x1, 5x7, 11x13, 17x19, 23x25, ...
$endgroup$
– OHO
yesterday
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
yesterday
$begingroup$
Or Maybe there is a Shorter way that im missing
$endgroup$
– Bedi
yesterday
$begingroup$
Or Maybe there is a Shorter way that im missing
$endgroup$
– Bedi
yesterday
$begingroup$
My comment might be confusing after the edit so I removed it, but +1 from me now
$endgroup$
– PunPun1000
yesterday
$begingroup$
My comment might be confusing after the edit so I removed it, but +1 from me now
$endgroup$
– PunPun1000
yesterday
1
1
$begingroup$
Or (-1)x1, 5x7, 11x13, 17x19, 23x25, ...
$endgroup$
– OHO
yesterday
$begingroup$
Or (-1)x1, 5x7, 11x13, 17x19, 23x25, ...
$endgroup$
– OHO
yesterday
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
yesterday
$begingroup$
@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
yesterday
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
answered yesterday
arbitrahjarbitrahj
630111
$endgroup$
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
answered yesterday
arbitrahjarbitrahj
630111
$endgroup$
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
answered yesterday
arbitrahjarbitrahj
630111
$endgroup$
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 yesterday
arbitrahjarbitrahj
630111
answered yesterday
arbitrahjarbitrahj
630111
answered yesterday
arbitrahjarbitrahj
630111
answered yesterday
arbitrahjarbitrahj
630111
630111
add a comment |
add a comment |
T. Arunkumar is a new contributor. Be nice, and check out our Code of Conduct.
T. Arunkumar is a new contributor. Be nice, and check out our Code of Conduct.
T. Arunkumar is a new contributor. Be nice, and check out our Code of Conduct.
T. Arunkumar is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Hm.. Maybe the first number should be $6$ instead of $-1$?
$endgroup$
– athin
yesterday
2
$begingroup$
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
yesterday