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Spiral Stumper Series: Instructionless Puzzle

Find the number for the question mark

IQ-test-type question with not satisfying answerWhich number should replace the question mark?Find the missing piece to complete the patternCan you find the value of the question mark$colorred?$Reasoning - Find Missing NumberWhich two numbers replaces the question mark?Can you find the next number in the following series?What is the missing number in the question mark?What is the missing number, can anyone solve this? My original puzzleWhat is the pattern in the question box?

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;

.everyonelovesstackoverflowposition:absolute;height:1px;width:1px;opacity:0;top:0;left:0;pointer-events:none;

2

$begingroup$

What is the number for the question mark? (My original puzzle)

3=39

1=3

4=84

5=155

2=14

10=?

pattern

share|improve this question

asked 8 hours ago

Deepthinker101Deepthinker101

1961010 bronze badges

$endgroup$

add a comment
 | 

2

$begingroup$

What is the number for the question mark? (My original puzzle)

3=39

1=3

4=84

5=155

2=14

10=?

pattern

share|improve this question

asked 8 hours ago

Deepthinker101Deepthinker101

1961010 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

What is the number for the question mark? (My original puzzle)

3=39

1=3

4=84

5=155

2=14

10=?

pattern

share|improve this question

asked 8 hours ago

Deepthinker101Deepthinker101

1961010 bronze badges

$endgroup$

share|improve this question

asked 8 hours ago

Deepthinker101Deepthinker101

1961010 bronze badges

share|improve this question

asked 8 hours ago

Deepthinker101Deepthinker101

1961010 bronze badges

asked 8 hours ago

Deepthinker101Deepthinker101

1961010 bronze badges

asked 8 hours ago

Deepthinker101Deepthinker101

1961010 bronze badges

2 Answers
2

active

oldest

votes

2

$begingroup$

By noting that each number maps to a multiple of itself:
$3rightarrow3*13$
$1rightarrow1*3$
$4rightarrow4*21$
$5rightarrow5*31$
$2rightarrow2*7$

And that the quotient increases quadratically
$1:3$
$2:7=3+4$
$3:13=7+6$
$4:21=13+8$
$5:31=21+10$

The mapping must be a cubic function.

The above sequence is equal to $n^2+n+1$, where n is the input.

The original sequence is multiplied by $n$ again.
So the function can be described by $n^3+n^2+n$.

So the final answer is $10^3+10^2+10=1110$.

share|improve this answer

answered 1 hour ago

Matthew JensenMatthew Jensen

31144 bronze badges

New contributor

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

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It looks like this is it!
  $endgroup$
  – Duck
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Congratulations
  $endgroup$
  – Deepthinker101
  14 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This was the mathematical way to do what I described in words. Congratulations!
  $endgroup$
  – El-Guest
  7 mins ago
  

  
  
  
  

add a comment
 | 

4

$begingroup$

Note that

1 x 3 = 3 (3 is the 2nd odd positive number)

2 x 7 = 14 (7 is the 4th positive odd number)

3 x 13 = 39 (13 is the 7th positive odd number)

4 x 21 = 84 (21 is the 11th positive odd number)

5 x 31 = 155 (31 is the 16th positive odd number)

Note that

The nth odd number is being used, where n forms a pattern of 2, 4, 7, 11, 16 — ie. differences increase by one. Then to continue this pattern, we have 22, 29, 37, 46, 56 — this sequence is the triangular numbers plus 1. (Note the triangular Numbers are 1, 3, 6, 10, 15, 21, 28, 36, 45, 55...)

So then the answer ought to be

10 = 1110, as 111 is the 56th positive odd number and 10 x 111 = 1110.

share|improve this answer

edited 8 mins ago

answered 8 hours ago

El-GuestEl-Guest

26.4k33 gold badges6363 silver badges108108 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are probably on the right track, but maybe insisting on primes isn't necessary?
  $endgroup$
  – Bass
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, it looks like you’re right — this fits better I think. Thanks for the heads up, @Bass
  $endgroup$
  – El-Guest
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No not the right answer
  $endgroup$
  – Deepthinker101
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As it turns out, my process IS right, I’m just a dumdum.
  $endgroup$
  – El-Guest
  8 mins ago
  

  
  
  
  

add a comment
 | 

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2

$begingroup$

By noting that each number maps to a multiple of itself:
$3rightarrow3*13$
$1rightarrow1*3$
$4rightarrow4*21$
$5rightarrow5*31$
$2rightarrow2*7$

And that the quotient increases quadratically
$1:3$
$2:7=3+4$
$3:13=7+6$
$4:21=13+8$
$5:31=21+10$

The mapping must be a cubic function.

The above sequence is equal to $n^2+n+1$, where n is the input.

The original sequence is multiplied by $n$ again.
So the function can be described by $n^3+n^2+n$.

So the final answer is $10^3+10^2+10=1110$.

share|improve this answer

answered 1 hour ago

Matthew JensenMatthew Jensen

31144 bronze badges

New contributor

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

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It looks like this is it!
  $endgroup$
  – Duck
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Congratulations
  $endgroup$
  – Deepthinker101
  14 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This was the mathematical way to do what I described in words. Congratulations!
  $endgroup$
  – El-Guest
  7 mins ago
  

  
  
  
  

add a comment
 | 

2

$begingroup$

By noting that each number maps to a multiple of itself:
$3rightarrow3*13$
$1rightarrow1*3$
$4rightarrow4*21$
$5rightarrow5*31$
$2rightarrow2*7$

And that the quotient increases quadratically
$1:3$
$2:7=3+4$
$3:13=7+6$
$4:21=13+8$
$5:31=21+10$

The mapping must be a cubic function.

The above sequence is equal to $n^2+n+1$, where n is the input.

The original sequence is multiplied by $n$ again.
So the function can be described by $n^3+n^2+n$.

So the final answer is $10^3+10^2+10=1110$.

share|improve this answer

answered 1 hour ago

Matthew JensenMatthew Jensen

31144 bronze badges

New contributor

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

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It looks like this is it!
  $endgroup$
  – Duck
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Congratulations
  $endgroup$
  – Deepthinker101
  14 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This was the mathematical way to do what I described in words. Congratulations!
  $endgroup$
  – El-Guest
  7 mins ago
  

  
  
  
  

add a comment
 | 

$begingroup$

By noting that each number maps to a multiple of itself:
$3rightarrow3*13$
$1rightarrow1*3$
$4rightarrow4*21$
$5rightarrow5*31$
$2rightarrow2*7$

And that the quotient increases quadratically
$1:3$
$2:7=3+4$
$3:13=7+6$
$4:21=13+8$
$5:31=21+10$

The mapping must be a cubic function.

The above sequence is equal to $n^2+n+1$, where n is the input.

The original sequence is multiplied by $n$ again.
So the function can be described by $n^3+n^2+n$.

So the final answer is $10^3+10^2+10=1110$.

share|improve this answer

answered 1 hour ago

Matthew JensenMatthew Jensen

31144 bronze badges

New contributor

Matthew Jensen 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

answered 1 hour ago

Matthew JensenMatthew Jensen

31144 bronze badges

New contributor

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

answered 1 hour ago

Matthew JensenMatthew Jensen

31144 bronze badges

New contributor

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

answered 1 hour ago

Matthew JensenMatthew Jensen

31144 bronze badges

4

$begingroup$

Note that

1 x 3 = 3 (3 is the 2nd odd positive number)

2 x 7 = 14 (7 is the 4th positive odd number)

3 x 13 = 39 (13 is the 7th positive odd number)

4 x 21 = 84 (21 is the 11th positive odd number)

5 x 31 = 155 (31 is the 16th positive odd number)

Note that

The nth odd number is being used, where n forms a pattern of 2, 4, 7, 11, 16 — ie. differences increase by one. Then to continue this pattern, we have 22, 29, 37, 46, 56 — this sequence is the triangular numbers plus 1. (Note the triangular Numbers are 1, 3, 6, 10, 15, 21, 28, 36, 45, 55...)

So then the answer ought to be

10 = 1110, as 111 is the 56th positive odd number and 10 x 111 = 1110.

share|improve this answer

edited 8 mins ago

answered 8 hours ago

El-GuestEl-Guest

26.4k33 gold badges6363 silver badges108108 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are probably on the right track, but maybe insisting on primes isn't necessary?
  $endgroup$
  – Bass
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, it looks like you’re right — this fits better I think. Thanks for the heads up, @Bass
  $endgroup$
  – El-Guest
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No not the right answer
  $endgroup$
  – Deepthinker101
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As it turns out, my process IS right, I’m just a dumdum.
  $endgroup$
  – El-Guest
  8 mins ago
  

  
  
  
  

add a comment
 | 

4

$begingroup$

Note that

1 x 3 = 3 (3 is the 2nd odd positive number)

2 x 7 = 14 (7 is the 4th positive odd number)

3 x 13 = 39 (13 is the 7th positive odd number)

4 x 21 = 84 (21 is the 11th positive odd number)

5 x 31 = 155 (31 is the 16th positive odd number)

Note that

The nth odd number is being used, where n forms a pattern of 2, 4, 7, 11, 16 — ie. differences increase by one. Then to continue this pattern, we have 22, 29, 37, 46, 56 — this sequence is the triangular numbers plus 1. (Note the triangular Numbers are 1, 3, 6, 10, 15, 21, 28, 36, 45, 55...)

So then the answer ought to be

10 = 1110, as 111 is the 56th positive odd number and 10 x 111 = 1110.

share|improve this answer

edited 8 mins ago

answered 8 hours ago

El-GuestEl-Guest

26.4k33 gold badges6363 silver badges108108 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are probably on the right track, but maybe insisting on primes isn't necessary?
  $endgroup$
  – Bass
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, it looks like you’re right — this fits better I think. Thanks for the heads up, @Bass
  $endgroup$
  – El-Guest
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No not the right answer
  $endgroup$
  – Deepthinker101
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As it turns out, my process IS right, I’m just a dumdum.
  $endgroup$
  – El-Guest
  8 mins ago
  

  
  
  
  

add a comment
 | 

$begingroup$

Note that

1 x 3 = 3 (3 is the 2nd odd positive number)

2 x 7 = 14 (7 is the 4th positive odd number)

3 x 13 = 39 (13 is the 7th positive odd number)

4 x 21 = 84 (21 is the 11th positive odd number)

5 x 31 = 155 (31 is the 16th positive odd number)

Note that

The nth odd number is being used, where n forms a pattern of 2, 4, 7, 11, 16 — ie. differences increase by one. Then to continue this pattern, we have 22, 29, 37, 46, 56 — this sequence is the triangular numbers plus 1. (Note the triangular Numbers are 1, 3, 6, 10, 15, 21, 28, 36, 45, 55...)

So then the answer ought to be

10 = 1110, as 111 is the 56th positive odd number and 10 x 111 = 1110.

share|improve this answer

edited 8 mins ago

answered 8 hours ago

El-GuestEl-Guest

26.4k33 gold badges6363 silver badges108108 bronze badges

$endgroup$

share|improve this answer

edited 8 mins ago

answered 8 hours ago

El-GuestEl-Guest

26.4k33 gold badges6363 silver badges108108 bronze badges

answered 8 hours ago

El-GuestEl-Guest

26.4k33 gold badges6363 silver badges108108 bronze badges

answered 8 hours ago

El-GuestEl-Guest

26.4k33 gold badges6363 silver badges108108 bronze badges