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A Non Math Puzzle. What is the middle number?

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

18

1

$begingroup$

There are three patterns below.

Two diagonal and one vertical. The middle number is common to them all

What is it?

It is not mathematical per se.

NO partial answers. Explain all three patterns in your answer.

pattern lateral-thinking

share|improve this question

asked Aug 12 at 13:19

DEEMDEEM

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$endgroup$

add a comment | 

18

1

$begingroup$

There are three patterns below.

Two diagonal and one vertical. The middle number is common to them all

What is it?

It is not mathematical per se.

NO partial answers. Explain all three patterns in your answer.

pattern lateral-thinking

share|improve this question

asked Aug 12 at 13:19

DEEMDEEM

8,10111 gold badge2626 silver badges142142 bronze badges

$endgroup$

add a comment | 

$begingroup$

There are three patterns below.

Two diagonal and one vertical. The middle number is common to them all

What is it?

It is not mathematical per se.

NO partial answers. Explain all three patterns in your answer.

pattern lateral-thinking

share|improve this question

asked Aug 12 at 13:19

DEEMDEEM

8,10111 gold badge2626 silver badges142142 bronze badges

$endgroup$

Two diagonal and one vertical. The middle number is common to them all

What is it?

share|improve this question

asked Aug 12 at 13:19

DEEMDEEM

8,10111 gold badge2626 silver badges142142 bronze badges

share|improve this question

asked Aug 12 at 13:19

DEEMDEEM

8,10111 gold badge2626 silver badges142142 bronze badges

asked Aug 12 at 13:19

DEEMDEEM

8,10111 gold badge2626 silver badges142142 bronze badges

asked Aug 12 at 13:19

DEEMDEEM

8,10111 gold badge2626 silver badges142142 bronze badges

2 Answers
2

active

oldest

votes

24

$begingroup$

I think the middle number is

$11$

First sequence

$2,6,11,51,101$

These are one more than the first five Roman numerals with one letter $(I, V, X, L, C)$.

Second sequence

$1,8,11,18,80$

These are the first five numbers which begin with a vowel in English.

Third sequence

$1,3,11,17,111$

When written in English the $n$th term of this sequence is the smallest which contains $n$ Es.

share|improve this answer

answered Aug 12 at 14:03

hexominohexomino

59.9k55 gold badges172172 silver badges271271 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Kudos @hexomino. Special for you. What are the next three numbers after 1,3,11,17,111 assuming 4-6 digit numbers are written as _ thousand etc.
  $endgroup$
  – DEEM
  Aug 12 at 15:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @DEEM I guess they would be 117, 317 and 1317. There's a little bit of a jump here but we can deduce from the previous terms in the sequence that there will be no term after 317 less than 1000 and adding 1000 just gives one extra E.
  $endgroup$
  – hexomino
  Aug 12 at 15:36
  
  

  
  
  
  

add a comment | 

3

$begingroup$

The quiz is very open to interpretation, so this is mine.

Every line sets a a lower bound for the numbers that can appear on the next. In the 1st line, the biggest number is 2, meaning that the second line contains only numbers > 2. The same logic applies in reverse, so 4th line smallest number is 17, hence, 3rd line must contain a number < 17.

Applying this rule, 8 > ? > 17. Suppose the aim of the game was to have all the digits in the highest number of each sequence (namely, 111 - 80 - 101) previously mentioned in the sequence itself.

111 can take the 1s from 1 and 17 but it's still missing one 1;

80 can take the 8 from either 8 or 18 but it's still missing the 0;

101 can take the 1 from 51 but it's missing both the 0 and one more 1.

This leads to the final solution:

10

share|improve this answer

edited Aug 13 at 13:04

gabbo1092

4,95588 silver badges3939 bronze badges

answered Aug 13 at 10:20

Alessandro AmantiniAlessandro Amantini

3111 bronze badge

$endgroup$

add a comment | 

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24

$begingroup$

I think the middle number is

$11$

First sequence

$2,6,11,51,101$

These are one more than the first five Roman numerals with one letter $(I, V, X, L, C)$.

Second sequence

$1,8,11,18,80$

These are the first five numbers which begin with a vowel in English.

Third sequence

$1,3,11,17,111$

When written in English the $n$th term of this sequence is the smallest which contains $n$ Es.

share|improve this answer

answered Aug 12 at 14:03

hexominohexomino

59.9k55 gold badges172172 silver badges271271 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Kudos @hexomino. Special for you. What are the next three numbers after 1,3,11,17,111 assuming 4-6 digit numbers are written as _ thousand etc.
  $endgroup$
  – DEEM
  Aug 12 at 15:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @DEEM I guess they would be 117, 317 and 1317. There's a little bit of a jump here but we can deduce from the previous terms in the sequence that there will be no term after 317 less than 1000 and adding 1000 just gives one extra E.
  $endgroup$
  – hexomino
  Aug 12 at 15:36
  
  

  
  
  
  

add a comment | 

24

$begingroup$

I think the middle number is

$11$

First sequence

$2,6,11,51,101$

These are one more than the first five Roman numerals with one letter $(I, V, X, L, C)$.

Second sequence

$1,8,11,18,80$

These are the first five numbers which begin with a vowel in English.

Third sequence

$1,3,11,17,111$

When written in English the $n$th term of this sequence is the smallest which contains $n$ Es.

share|improve this answer

answered Aug 12 at 14:03

hexominohexomino

59.9k55 gold badges172172 silver badges271271 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Kudos @hexomino. Special for you. What are the next three numbers after 1,3,11,17,111 assuming 4-6 digit numbers are written as _ thousand etc.
  $endgroup$
  – DEEM
  Aug 12 at 15:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @DEEM I guess they would be 117, 317 and 1317. There's a little bit of a jump here but we can deduce from the previous terms in the sequence that there will be no term after 317 less than 1000 and adding 1000 just gives one extra E.
  $endgroup$
  – hexomino
  Aug 12 at 15:36
  
  

  
  
  
  

add a comment | 

$begingroup$

I think the middle number is

$11$

First sequence

$2,6,11,51,101$

These are one more than the first five Roman numerals with one letter $(I, V, X, L, C)$.

Second sequence

$1,8,11,18,80$

These are the first five numbers which begin with a vowel in English.

Third sequence

$1,3,11,17,111$

When written in English the $n$th term of this sequence is the smallest which contains $n$ Es.

share|improve this answer

answered Aug 12 at 14:03

hexominohexomino

59.9k55 gold badges172172 silver badges271271 bronze badges

$endgroup$

share|improve this answer

answered Aug 12 at 14:03

hexominohexomino

59.9k55 gold badges172172 silver badges271271 bronze badges

answered Aug 12 at 14:03

hexominohexomino

59.9k55 gold badges172172 silver badges271271 bronze badges

answered Aug 12 at 14:03

hexominohexomino

59.9k55 gold badges172172 silver badges271271 bronze badges

3

$begingroup$

The quiz is very open to interpretation, so this is mine.

Every line sets a a lower bound for the numbers that can appear on the next. In the 1st line, the biggest number is 2, meaning that the second line contains only numbers > 2. The same logic applies in reverse, so 4th line smallest number is 17, hence, 3rd line must contain a number < 17.

Applying this rule, 8 > ? > 17. Suppose the aim of the game was to have all the digits in the highest number of each sequence (namely, 111 - 80 - 101) previously mentioned in the sequence itself.

111 can take the 1s from 1 and 17 but it's still missing one 1;

80 can take the 8 from either 8 or 18 but it's still missing the 0;

101 can take the 1 from 51 but it's missing both the 0 and one more 1.

This leads to the final solution:

10

share|improve this answer

edited Aug 13 at 13:04

gabbo1092

4,95588 silver badges3939 bronze badges

answered Aug 13 at 10:20

Alessandro AmantiniAlessandro Amantini

3111 bronze badge

$endgroup$

add a comment | 

3

$begingroup$

The quiz is very open to interpretation, so this is mine.

Every line sets a a lower bound for the numbers that can appear on the next. In the 1st line, the biggest number is 2, meaning that the second line contains only numbers > 2. The same logic applies in reverse, so 4th line smallest number is 17, hence, 3rd line must contain a number < 17.

Applying this rule, 8 > ? > 17. Suppose the aim of the game was to have all the digits in the highest number of each sequence (namely, 111 - 80 - 101) previously mentioned in the sequence itself.

111 can take the 1s from 1 and 17 but it's still missing one 1;

80 can take the 8 from either 8 or 18 but it's still missing the 0;

101 can take the 1 from 51 but it's missing both the 0 and one more 1.

This leads to the final solution:

10

share|improve this answer

edited Aug 13 at 13:04

gabbo1092

4,95588 silver badges3939 bronze badges

answered Aug 13 at 10:20

Alessandro AmantiniAlessandro Amantini

3111 bronze badge

$endgroup$

add a comment | 

$begingroup$

The quiz is very open to interpretation, so this is mine.

Every line sets a a lower bound for the numbers that can appear on the next. In the 1st line, the biggest number is 2, meaning that the second line contains only numbers > 2. The same logic applies in reverse, so 4th line smallest number is 17, hence, 3rd line must contain a number < 17.

Applying this rule, 8 > ? > 17. Suppose the aim of the game was to have all the digits in the highest number of each sequence (namely, 111 - 80 - 101) previously mentioned in the sequence itself.

111 can take the 1s from 1 and 17 but it's still missing one 1;

80 can take the 8 from either 8 or 18 but it's still missing the 0;

101 can take the 1 from 51 but it's missing both the 0 and one more 1.

This leads to the final solution:

10

share|improve this answer

edited Aug 13 at 13:04

gabbo1092

4,95588 silver badges3939 bronze badges

answered Aug 13 at 10:20

Alessandro AmantiniAlessandro Amantini

3111 bronze badge

$endgroup$

share|improve this answer

edited Aug 13 at 13:04

gabbo1092

4,95588 silver badges3939 bronze badges

answered Aug 13 at 10:20

Alessandro AmantiniAlessandro Amantini

3111 bronze badge

edited Aug 13 at 13:04

gabbo1092

4,95588 silver badges3939 bronze badges

edited Aug 13 at 13:04

gabbo1092

4,95588 silver badges3939 bronze badges

answered Aug 13 at 10:20

Alessandro AmantiniAlessandro Amantini

3111 bronze badge

answered Aug 13 at 10:20

Alessandro AmantiniAlessandro Amantini

3111 bronze badge