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A Non Math Puzzle. What is the middle number?
Find the next image in this abstract reasoning puzzleUnconventional Math - Anti-Number PuzzleSticking NumbersIllegal patterns for the Rubik's cubeCannot see the patternThey are all tied togetherWhat are the house numbers?Ask one verbal question to figure out who is blind and who is mute among three persons
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$begingroup$
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
asked Aug 12 at 13:19
DEEMDEEM
17.1k1 gold badge32 silver badges158 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
asked Aug 12 at 13:19
DEEMDEEM
17.1k1 gold badge32 silver badges158 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
asked Aug 12 at 13:19
DEEMDEEM
17.1k1 gold badge32 silver badges158 bronze badges
$endgroup$
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
pattern lateral-thinking
asked Aug 12 at 13:19
DEEMDEEM
17.1k1 gold badge32 silver badges158 bronze badges
asked Aug 12 at 13:19
DEEMDEEM
17.1k1 gold badge32 silver badges158 bronze badges
asked Aug 12 at 13:19
DEEMDEEM
17.1k1 gold badge32 silver badges158 bronze badges
asked Aug 12 at 13:19
DEEMDEEM
17.1k1 gold badge32 silver badges158 bronze badges
asked Aug 12 at 13:19
DEEMDEEM
17.1k1 gold badge32 silver badges158 bronze badges
17.1k1 gold badge32 silver badges158 bronze badges
add a comment
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2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Aug 12 at 14:03
hexominohexomino
69k6 gold badges194 silver badges298 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
|
$begingroup$
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
edited Aug 13 at 13:04
gabbo1092
5,3409 silver badges39 bronze badges
answered Aug 13 at 10:20
Alessandro AmantiniAlessandro Amantini
311 bronze badge
$endgroup$
add a comment
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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 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.
answered Aug 12 at 14:03
hexominohexomino
69k6 gold badges194 silver badges298 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.
answered Aug 12 at 14:03
hexominohexomino
69k6 gold badges194 silver badges298 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.
answered Aug 12 at 14:03
hexominohexomino
69k6 gold badges194 silver badges298 bronze badges
$endgroup$
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.
answered Aug 12 at 14:03
hexominohexomino
69k6 gold badges194 silver badges298 bronze badges
answered Aug 12 at 14:03
hexominohexomino
69k6 gold badges194 silver badges298 bronze badges
answered Aug 12 at 14:03
hexominohexomino
69k6 gold badges194 silver badges298 bronze badges
answered Aug 12 at 14:03
hexominohexomino
69k6 gold badges194 silver badges298 bronze badges
69k6 gold badges194 silver badges298 bronze badges
$begingroup$
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
|
$begingroup$
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
$begingroup$
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
$begingroup$
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
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
$begingroup$
@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
|
$begingroup$
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
edited Aug 13 at 13:04
gabbo1092
5,3409 silver badges39 bronze badges
answered Aug 13 at 10:20
Alessandro AmantiniAlessandro Amantini
311 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
edited Aug 13 at 13:04
gabbo1092
5,3409 silver badges39 bronze badges
answered Aug 13 at 10:20
Alessandro AmantiniAlessandro Amantini
311 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
edited Aug 13 at 13:04
gabbo1092
5,3409 silver badges39 bronze badges
answered Aug 13 at 10:20
Alessandro AmantiniAlessandro Amantini
311 bronze badge
$endgroup$
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
edited Aug 13 at 13:04
gabbo1092
5,3409 silver badges39 bronze badges
answered Aug 13 at 10:20
Alessandro AmantiniAlessandro Amantini
311 bronze badge
edited Aug 13 at 13:04
gabbo1092
5,3409 silver badges39 bronze badges
edited Aug 13 at 13:04
gabbo1092
5,3409 silver badges39 bronze badges
edited Aug 13 at 13:04
gabbo1092
5,3409 silver badges39 bronze badges
5,3409 silver badges39 bronze badges
answered Aug 13 at 10:20
Alessandro AmantiniAlessandro Amantini
311 bronze badge
answered Aug 13 at 10:20
Alessandro AmantiniAlessandro Amantini
311 bronze badge
answered Aug 13 at 10:20
Alessandro AmantiniAlessandro Amantini
311 bronze badge
311 bronze badge
add a comment
|
add a comment
|
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