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What is the question mark?

What is the next number in sequence?Which number should replace the question mark?Which number will replace the question markThe next number in sequenceWhich two numbers replaces the question mark?Which number replaces the question mark in the circle?Number Sequence Series – Question 1Number Sequence Series-Question 2Number Sequence Series-Question 3Number Sequence Series-Question 4

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2

1

$begingroup$

My friend gave me this sequence of numbers：

---

3=11

---

6=12

---

11=13

---

14=14

---

361114=A

---

A=?

---

Which number can replace the question mark?

number-sequence

share|improve this question

edited Aug 16 at 9:27

Tanton

asked Aug 16 at 9:20

TantonTanton

3544 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is the original!
  $endgroup$
  – Tanton
  Aug 16 at 9:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does A change the answer?
  $endgroup$
  – Abbas
  Aug 16 at 9:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A It doesn't make any sense
  $endgroup$
  – Tanton
  Aug 16 at 10:34
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Could the last two lines have been replaced with 361114=? instead of adding a variable into the mix?
  $endgroup$
  – Voldemort's Wrath
  Aug 16 at 13:17
  

  
  
  
  

add a comment | 

2

1

$begingroup$

My friend gave me this sequence of numbers：

---

3=11

---

6=12

---

11=13

---

14=14

---

361114=A

---

A=?

---

Which number can replace the question mark?

number-sequence

share|improve this question

edited Aug 16 at 9:27

Tanton

asked Aug 16 at 9:20

TantonTanton

3544 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is the original!
  $endgroup$
  – Tanton
  Aug 16 at 9:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does A change the answer?
  $endgroup$
  – Abbas
  Aug 16 at 9:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A It doesn't make any sense
  $endgroup$
  – Tanton
  Aug 16 at 10:34
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Could the last two lines have been replaced with 361114=? instead of adding a variable into the mix?
  $endgroup$
  – Voldemort's Wrath
  Aug 16 at 13:17
  

  
  
  
  

add a comment | 

$begingroup$

My friend gave me this sequence of numbers：

---

3=11

---

6=12

---

11=13

---

14=14

---

361114=A

---

A=?

---

Which number can replace the question mark?

number-sequence

share|improve this question

edited Aug 16 at 9:27

Tanton

asked Aug 16 at 9:20

TantonTanton

3544 bronze badges

$endgroup$

share|improve this question

edited Aug 16 at 9:27

Tanton

asked Aug 16 at 9:20

TantonTanton

3544 bronze badges

share|improve this question

edited Aug 16 at 9:27

Tanton

asked Aug 16 at 9:20

TantonTanton

3544 bronze badges

asked Aug 16 at 9:20

TantonTanton

3544 bronze badges

asked Aug 16 at 9:20

TantonTanton

3544 bronze badges

4 Answers
4

active

oldest

votes

3

$begingroup$

New Solution:

10. A (hexadecimal) = 10

Another possible Solution could be:

1. Since A is the first letter in the alphabet

If it's

11121314

then it's too easy.

share|improve this answer

edited Aug 17 at 8:50

answered Aug 16 at 9:22

AbbasAbbas

49488 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A is just a medium for transformation!
  $endgroup$
  – Tanton
  Aug 16 at 9:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm thinking the last one here is the most applicable, and probably the answer to the question. Can you confirm/deny @Tanton?
  $endgroup$
  – Birjolaxew
  Aug 16 at 10:41
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  @Tanton maybe you should discuss this with your friend before you accept this answer
  $endgroup$
  – Adam
  Aug 16 at 11:14
  

  
  
  
  


• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  $begingroup$
  This answer currently says that "A (hexadecimal) = 11" but in hexadecimal A means 10. No part of it has any actual explanation that I can make any sense of it. @Tanton I really don't think this should be accepted.
  $endgroup$
  – Gareth McCaughan♦
  Aug 16 at 21:23
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I literally cannot understand what you're saying with this answer. Both "new" and "another possible" solutions ignore the entirety of the puzzle save its final entry and focus on the letter, whose specific inclusion may indeed be relevant somehow but certainly shouldn't be the be-all-end-all of the puzzle. This is supposed to be a sequence of numbers; you've discarded the sequence entirely and just proposed a substitution of number for variable. How is this a solution to the sequence? @Tanton how is this possibly a satisfying answer?
  $endgroup$
  – Rubio♦
  Aug 17 at 15:37
  

  
  
  
  

 | 
show 8 more comments

5

$begingroup$

I also have a solution

123468

because,

Think of the number to the left of the sequence as octal

Like

3（octal）= 11（binary）
6（octal）= 12（quanternary）
11（octal）= 13（Senary）
14（octal）= 14（octal）

So

361114（octal）= 123468（decimal）

share|improve this answer

edited Aug 16 at 14:44

Adam

93611 gold badge22 silver badges2626 bronze badges

answered Aug 16 at 14:20

sypickysypicky

6955 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My answer is based on the fact that A has no effect. Because A=123468 is substituted into the next formula, I can't continue to convert.
  $endgroup$
  – sypicky
  Aug 16 at 14:41
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What determines the bases used on the right-hand sides?
  $endgroup$
  – Gareth McCaughan♦
  Aug 16 at 21:25
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  On the right hand side is a series of even numbers, and adds 2.I think it's a rule.
  $endgroup$
  – sypicky
  Aug 17 at 0:05
  

  
  
  
  

add a comment | 

4

$begingroup$

An alternate solution is:

24024

because,

generally, when you have variable names next to each other (e.g. $xy$) that would mean that you are multiplying their values (e.g. $xtimes y$).

Therefore,

you would multiply $11times12times13times14=24024$

share|improve this answer

edited Aug 16 at 13:13

answered Aug 16 at 12:46

Voldemort's WrathVoldemort's Wrath

44511 silver badge2323 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Also a valid theory, It could be your answer is the one OP's friend was looking for!
  $endgroup$
  – Abbas
  Aug 16 at 14:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How does this account for things like "11=13" in the question?
  $endgroup$
  – Gareth McCaughan♦
  Aug 16 at 21:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe I'm just slow today... why did you select those numbers to multiply?
  $endgroup$
  – maxathousand
  Aug 16 at 21:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't think multipliers can be omitted between numbers. One example is 6÷2 (1 + 2).
  $endgroup$
  – sypicky
  Aug 17 at 0:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @sypicky - That's true, but neither can you say $3=11$... This is assuming the numbers on the left are variables.
  $endgroup$
  – Voldemort's Wrath
  Aug 17 at 1:46
  

  
  
  
  

 | 
show 4 more comments

0

$begingroup$

Based on the last but one line, A is

361114

share|improve this answer

answered Aug 16 at 16:39

Mea Culpa NayMea Culpa Nay

7,05411 gold badge77 silver badges4242 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well if A is this number then why is 3=11 or 6=12 or 11=13?
  $endgroup$
  – Adam
  Aug 16 at 18:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As, I have just seen Voldermort's 1st comment to the OP and up-voted it.
  $endgroup$
  – Mea Culpa Nay
  Aug 17 at 6:34
  
  

  
  
  
  

add a comment | 

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3

$begingroup$

New Solution:

10. A (hexadecimal) = 10

Another possible Solution could be:

1. Since A is the first letter in the alphabet

If it's

11121314

then it's too easy.

share|improve this answer

edited Aug 17 at 8:50

answered Aug 16 at 9:22

AbbasAbbas

49488 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A is just a medium for transformation!
  $endgroup$
  – Tanton
  Aug 16 at 9:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm thinking the last one here is the most applicable, and probably the answer to the question. Can you confirm/deny @Tanton?
  $endgroup$
  – Birjolaxew
  Aug 16 at 10:41
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  @Tanton maybe you should discuss this with your friend before you accept this answer
  $endgroup$
  – Adam
  Aug 16 at 11:14
  

  
  
  
  


• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  $begingroup$
  This answer currently says that "A (hexadecimal) = 11" but in hexadecimal A means 10. No part of it has any actual explanation that I can make any sense of it. @Tanton I really don't think this should be accepted.
  $endgroup$
  – Gareth McCaughan♦
  Aug 16 at 21:23
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I literally cannot understand what you're saying with this answer. Both "new" and "another possible" solutions ignore the entirety of the puzzle save its final entry and focus on the letter, whose specific inclusion may indeed be relevant somehow but certainly shouldn't be the be-all-end-all of the puzzle. This is supposed to be a sequence of numbers; you've discarded the sequence entirely and just proposed a substitution of number for variable. How is this a solution to the sequence? @Tanton how is this possibly a satisfying answer?
  $endgroup$
  – Rubio♦
  Aug 17 at 15:37
  

  
  
  
  

 | 
show 8 more comments

3

$begingroup$

New Solution:

10. A (hexadecimal) = 10

Another possible Solution could be:

1. Since A is the first letter in the alphabet

If it's

11121314

then it's too easy.

share|improve this answer

edited Aug 17 at 8:50

answered Aug 16 at 9:22

AbbasAbbas

49488 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A is just a medium for transformation!
  $endgroup$
  – Tanton
  Aug 16 at 9:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm thinking the last one here is the most applicable, and probably the answer to the question. Can you confirm/deny @Tanton?
  $endgroup$
  – Birjolaxew
  Aug 16 at 10:41
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  @Tanton maybe you should discuss this with your friend before you accept this answer
  $endgroup$
  – Adam
  Aug 16 at 11:14
  

  
  
  
  


• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  $begingroup$
  This answer currently says that "A (hexadecimal) = 11" but in hexadecimal A means 10. No part of it has any actual explanation that I can make any sense of it. @Tanton I really don't think this should be accepted.
  $endgroup$
  – Gareth McCaughan♦
  Aug 16 at 21:23
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I literally cannot understand what you're saying with this answer. Both "new" and "another possible" solutions ignore the entirety of the puzzle save its final entry and focus on the letter, whose specific inclusion may indeed be relevant somehow but certainly shouldn't be the be-all-end-all of the puzzle. This is supposed to be a sequence of numbers; you've discarded the sequence entirely and just proposed a substitution of number for variable. How is this a solution to the sequence? @Tanton how is this possibly a satisfying answer?
  $endgroup$
  – Rubio♦
  Aug 17 at 15:37
  

  
  
  
  

 | 
show 8 more comments

$begingroup$

New Solution:

10. A (hexadecimal) = 10

Another possible Solution could be:

1. Since A is the first letter in the alphabet

If it's

11121314

then it's too easy.

share|improve this answer

edited Aug 17 at 8:50

answered Aug 16 at 9:22

AbbasAbbas

49488 bronze badges

$endgroup$

share|improve this answer

edited Aug 17 at 8:50

answered Aug 16 at 9:22

AbbasAbbas

49488 bronze badges

answered Aug 16 at 9:22

AbbasAbbas

49488 bronze badges

answered Aug 16 at 9:22

AbbasAbbas

49488 bronze badges

5

$begingroup$

I also have a solution

123468

because,

Think of the number to the left of the sequence as octal

Like

3（octal）= 11（binary）
6（octal）= 12（quanternary）
11（octal）= 13（Senary）
14（octal）= 14（octal）

So

361114（octal）= 123468（decimal）

share|improve this answer

edited Aug 16 at 14:44

Adam

93611 gold badge22 silver badges2626 bronze badges

answered Aug 16 at 14:20

sypickysypicky

6955 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My answer is based on the fact that A has no effect. Because A=123468 is substituted into the next formula, I can't continue to convert.
  $endgroup$
  – sypicky
  Aug 16 at 14:41
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What determines the bases used on the right-hand sides?
  $endgroup$
  – Gareth McCaughan♦
  Aug 16 at 21:25
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  On the right hand side is a series of even numbers, and adds 2.I think it's a rule.
  $endgroup$
  – sypicky
  Aug 17 at 0:05
  

  
  
  
  

add a comment | 

5

$begingroup$

I also have a solution

123468

because,

Think of the number to the left of the sequence as octal

Like

3（octal）= 11（binary）
6（octal）= 12（quanternary）
11（octal）= 13（Senary）
14（octal）= 14（octal）

So

361114（octal）= 123468（decimal）

share|improve this answer

edited Aug 16 at 14:44

Adam

93611 gold badge22 silver badges2626 bronze badges

answered Aug 16 at 14:20

sypickysypicky

6955 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My answer is based on the fact that A has no effect. Because A=123468 is substituted into the next formula, I can't continue to convert.
  $endgroup$
  – sypicky
  Aug 16 at 14:41
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What determines the bases used on the right-hand sides?
  $endgroup$
  – Gareth McCaughan♦
  Aug 16 at 21:25
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  On the right hand side is a series of even numbers, and adds 2.I think it's a rule.
  $endgroup$
  – sypicky
  Aug 17 at 0:05
  

  
  
  
  

add a comment | 

$begingroup$

I also have a solution

123468

because,

Think of the number to the left of the sequence as octal

Like

3（octal）= 11（binary）
6（octal）= 12（quanternary）
11（octal）= 13（Senary）
14（octal）= 14（octal）

So

361114（octal）= 123468（decimal）

share|improve this answer

edited Aug 16 at 14:44

Adam

93611 gold badge22 silver badges2626 bronze badges

answered Aug 16 at 14:20

sypickysypicky

6955 bronze badges

$endgroup$

share|improve this answer

edited Aug 16 at 14:44

Adam

93611 gold badge22 silver badges2626 bronze badges

answered Aug 16 at 14:20

sypickysypicky

6955 bronze badges

edited Aug 16 at 14:44

Adam

93611 gold badge22 silver badges2626 bronze badges

edited Aug 16 at 14:44

Adam

93611 gold badge22 silver badges2626 bronze badges

answered Aug 16 at 14:20

sypickysypicky

6955 bronze badges

answered Aug 16 at 14:20

sypickysypicky

6955 bronze badges

4

$begingroup$

An alternate solution is:

24024

because,

generally, when you have variable names next to each other (e.g. $xy$) that would mean that you are multiplying their values (e.g. $xtimes y$).

Therefore,

you would multiply $11times12times13times14=24024$

share|improve this answer

edited Aug 16 at 13:13

answered Aug 16 at 12:46

Voldemort's WrathVoldemort's Wrath

44511 silver badge2323 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Also a valid theory, It could be your answer is the one OP's friend was looking for!
  $endgroup$
  – Abbas
  Aug 16 at 14:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How does this account for things like "11=13" in the question?
  $endgroup$
  – Gareth McCaughan♦
  Aug 16 at 21:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe I'm just slow today... why did you select those numbers to multiply?
  $endgroup$
  – maxathousand
  Aug 16 at 21:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't think multipliers can be omitted between numbers. One example is 6÷2 (1 + 2).
  $endgroup$
  – sypicky
  Aug 17 at 0:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @sypicky - That's true, but neither can you say $3=11$... This is assuming the numbers on the left are variables.
  $endgroup$
  – Voldemort's Wrath
  Aug 17 at 1:46
  

  
  
  
  

 | 
show 4 more comments

4

$begingroup$

An alternate solution is:

24024

because,

generally, when you have variable names next to each other (e.g. $xy$) that would mean that you are multiplying their values (e.g. $xtimes y$).

Therefore,

you would multiply $11times12times13times14=24024$

share|improve this answer

edited Aug 16 at 13:13

answered Aug 16 at 12:46

Voldemort's WrathVoldemort's Wrath

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  Also a valid theory, It could be your answer is the one OP's friend was looking for!
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  – Abbas
  Aug 16 at 14:15
  

  
  
  
  


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  How does this account for things like "11=13" in the question?
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  – Gareth McCaughan♦
  Aug 16 at 21:24
  

  
  
  
  


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  Maybe I'm just slow today... why did you select those numbers to multiply?
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  – maxathousand
  Aug 16 at 21:33
  

  
  
  
  


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  I don't think multipliers can be omitted between numbers. One example is 6÷2 (1 + 2).
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  – sypicky
  Aug 17 at 0:18
  

  
  
  
  


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  @sypicky - That's true, but neither can you say $3=11$... This is assuming the numbers on the left are variables.
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  – Voldemort's Wrath
  Aug 17 at 1:46
  

  
  
  
  

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An alternate solution is:

24024

because,

generally, when you have variable names next to each other (e.g. $xy$) that would mean that you are multiplying their values (e.g. $xtimes y$).

Therefore,

you would multiply $11times12times13times14=24024$

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edited Aug 16 at 13:13

answered Aug 16 at 12:46

Voldemort's WrathVoldemort's Wrath

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share|improve this answer

edited Aug 16 at 13:13

answered Aug 16 at 12:46

Voldemort's WrathVoldemort's Wrath

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answered Aug 16 at 12:46

Voldemort's WrathVoldemort's Wrath

44511 silver badge2323 bronze badges

answered Aug 16 at 12:46

Voldemort's WrathVoldemort's Wrath

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0

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Based on the last but one line, A is

361114

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answered Aug 16 at 16:39

Mea Culpa NayMea Culpa Nay

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  Well if A is this number then why is 3=11 or 6=12 or 11=13?
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  – Adam
  Aug 16 at 18:48
  

  
  
  
  


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  As, I have just seen Voldermort's 1st comment to the OP and up-voted it.
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  – Mea Culpa Nay
  Aug 17 at 6:34
  
  

  
  
  
  

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0

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Based on the last but one line, A is

361114

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answered Aug 16 at 16:39

Mea Culpa NayMea Culpa Nay

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  Well if A is this number then why is 3=11 or 6=12 or 11=13?
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  – Adam
  Aug 16 at 18:48
  

  
  
  
  


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  As, I have just seen Voldermort's 1st comment to the OP and up-voted it.
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  – Mea Culpa Nay
  Aug 17 at 6:34
  
  

  
  
  
  

add a comment | 

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Based on the last but one line, A is

361114

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answered Aug 16 at 16:39

Mea Culpa NayMea Culpa Nay

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share|improve this answer

answered Aug 16 at 16:39

Mea Culpa NayMea Culpa Nay

7,05411 gold badge77 silver badges4242 bronze badges

answered Aug 16 at 16:39

Mea Culpa NayMea Culpa Nay

7,05411 gold badge77 silver badges4242 bronze badges

answered Aug 16 at 16:39

Mea Culpa NayMea Culpa Nay

7,05411 gold badge77 silver badges4242 bronze badges