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Some numbers are more equivalent than others
Place the numbersSabotage at Sea - Cursed Cruise liner?How are the cows?Reading in the dark, faster than lightNumbers that could only growAnother sequence of numbersSome Really Confusing MathWhat are some good resources to practice logical puzzle-solving?Are these numbers unique?Which country has more?
$begingroup$
ALL ANIMALS ARE EQUAL
BUT SOME ANIMALS ARE MORE EQUAL THAN OTHERS
— from
Animal Farm
by George Orwell
A contrived simple
equivalence
rule applies neatly to numbers 0 through 99
but not to any other numbers.
Equivalences of numbers 0 through 19 are listed below,
accounting for almost all other eligible numbers as well,
where ‘=’ means “is equivalent to.”
(Each number is
reflexively
equivalent to itself.)
0 = no others
10 = no others
1 = no others
11 = 29 = 31 = 49 = 51 = 69 = 71 = 89 = 91
2 = no others
12 = 28 = 32 = 48 = 52 = 68 = 72 = 88 = 92
3 = no others
13 = 27 = 33 = 47 = 53 = 67 = 73 = 87 = 93
4 = no others
14 = 26 = 34 = 46 = 54 = 66 = 74 = 86 = 94
5 = no others
15 = 25 = 35 = 45 = 55 = 65 = 75 = 85 = 95
6 = no others
16 = 24 = 36 = 44 = 56 = 64 = 76 = 84 = 96
7 = no others
17 = 23 = 37 = 43 = 57 = 63 = 77 = 83 = 97
8 = no others
18 = 22 = 38 = 42 = 58 = 62 = 78 = 82 = 98
9 = no others
19 = 21 = 39 = 41 = 59 = 61 = 79 = 81 = 99
What would be the entry for 20 in this list?
20 = ___ . . . ?
Please use and explain the simplest possible rule,
not purely mathematical,
that accounts for every equivalence from 0 to 99.
lateral-thinking
asked 2 hours ago
humnhumn
14.6k442131
$endgroup$
add a comment |
$begingroup$
ALL ANIMALS ARE EQUAL
BUT SOME ANIMALS ARE MORE EQUAL THAN OTHERS
— from
Animal Farm
by George Orwell
A contrived simple
equivalence
rule applies neatly to numbers 0 through 99
but not to any other numbers.
Equivalences of numbers 0 through 19 are listed below,
accounting for almost all other eligible numbers as well,
where ‘=’ means “is equivalent to.”
(Each number is
reflexively
equivalent to itself.)
0 = no others
10 = no others
1 = no others
11 = 29 = 31 = 49 = 51 = 69 = 71 = 89 = 91
2 = no others
12 = 28 = 32 = 48 = 52 = 68 = 72 = 88 = 92
3 = no others
13 = 27 = 33 = 47 = 53 = 67 = 73 = 87 = 93
4 = no others
14 = 26 = 34 = 46 = 54 = 66 = 74 = 86 = 94
5 = no others
15 = 25 = 35 = 45 = 55 = 65 = 75 = 85 = 95
6 = no others
16 = 24 = 36 = 44 = 56 = 64 = 76 = 84 = 96
7 = no others
17 = 23 = 37 = 43 = 57 = 63 = 77 = 83 = 97
8 = no others
18 = 22 = 38 = 42 = 58 = 62 = 78 = 82 = 98
9 = no others
19 = 21 = 39 = 41 = 59 = 61 = 79 = 81 = 99
What would be the entry for 20 in this list?
20 = ___ . . . ?
Please use and explain the simplest possible rule,
not purely mathematical,
that accounts for every equivalence from 0 to 99.
lateral-thinking
asked 2 hours ago
humnhumn
14.6k442131
$endgroup$
1
$begingroup$
Apology for the lack of more specific tags: They would give away the solution.
$endgroup$
– humn
2 hours ago
3
$begingroup$
Hurray, a humn puzzle! It's been a while.
$endgroup$
– Rand al'Thor
1 hour ago
add a comment |
$begingroup$
ALL ANIMALS ARE EQUAL
BUT SOME ANIMALS ARE MORE EQUAL THAN OTHERS
— from
Animal Farm
by George Orwell
A contrived simple
equivalence
rule applies neatly to numbers 0 through 99
but not to any other numbers.
Equivalences of numbers 0 through 19 are listed below,
accounting for almost all other eligible numbers as well,
where ‘=’ means “is equivalent to.”
(Each number is
reflexively
equivalent to itself.)
0 = no others
10 = no others
1 = no others
11 = 29 = 31 = 49 = 51 = 69 = 71 = 89 = 91
2 = no others
12 = 28 = 32 = 48 = 52 = 68 = 72 = 88 = 92
3 = no others
13 = 27 = 33 = 47 = 53 = 67 = 73 = 87 = 93
4 = no others
14 = 26 = 34 = 46 = 54 = 66 = 74 = 86 = 94
5 = no others
15 = 25 = 35 = 45 = 55 = 65 = 75 = 85 = 95
6 = no others
16 = 24 = 36 = 44 = 56 = 64 = 76 = 84 = 96
7 = no others
17 = 23 = 37 = 43 = 57 = 63 = 77 = 83 = 97
8 = no others
18 = 22 = 38 = 42 = 58 = 62 = 78 = 82 = 98
9 = no others
19 = 21 = 39 = 41 = 59 = 61 = 79 = 81 = 99
What would be the entry for 20 in this list?
20 = ___ . . . ?
Please use and explain the simplest possible rule,
not purely mathematical,
that accounts for every equivalence from 0 to 99.
lateral-thinking
asked 2 hours ago
humnhumn
14.6k442131
$endgroup$
ALL ANIMALS ARE EQUAL
BUT SOME ANIMALS ARE MORE EQUAL THAN OTHERS
— from
Animal Farm
by George Orwell
A contrived simple
equivalence
rule applies neatly to numbers 0 through 99
but not to any other numbers.
Equivalences of numbers 0 through 19 are listed below,
accounting for almost all other eligible numbers as well,
where ‘=’ means “is equivalent to.”
(Each number is
reflexively
equivalent to itself.)
0 = no others
10 = no others
1 = no others
11 = 29 = 31 = 49 = 51 = 69 = 71 = 89 = 91
2 = no others
12 = 28 = 32 = 48 = 52 = 68 = 72 = 88 = 92
3 = no others
13 = 27 = 33 = 47 = 53 = 67 = 73 = 87 = 93
4 = no others
14 = 26 = 34 = 46 = 54 = 66 = 74 = 86 = 94
5 = no others
15 = 25 = 35 = 45 = 55 = 65 = 75 = 85 = 95
6 = no others
16 = 24 = 36 = 44 = 56 = 64 = 76 = 84 = 96
7 = no others
17 = 23 = 37 = 43 = 57 = 63 = 77 = 83 = 97
8 = no others
18 = 22 = 38 = 42 = 58 = 62 = 78 = 82 = 98
9 = no others
19 = 21 = 39 = 41 = 59 = 61 = 79 = 81 = 99
What would be the entry for 20 in this list?
20 = ___ . . . ?
Please use and explain the simplest possible rule,
not purely mathematical,
that accounts for every equivalence from 0 to 99.
lateral-thinking
lateral-thinking
asked 2 hours ago
humnhumn
14.6k442131
asked 2 hours ago
humnhumn
14.6k442131
asked 2 hours ago
humnhumn
14.6k442131
asked 2 hours ago
humnhumn
14.6k442131
asked 2 hours ago
humnhumn
14.6k442131
14.6k442131
1
$begingroup$
Apology for the lack of more specific tags: They would give away the solution.
$endgroup$
– humn
2 hours ago
3
$begingroup$
Hurray, a humn puzzle! It's been a while.
$endgroup$
– Rand al'Thor
1 hour ago
add a comment |
1
$begingroup$
Apology for the lack of more specific tags: They would give away the solution.
$endgroup$
– humn
2 hours ago
3
$begingroup$
Hurray, a humn puzzle! It's been a while.
$endgroup$
– Rand al'Thor
1 hour ago
1
1
$begingroup$
Apology for the lack of more specific tags: They would give away the solution.
$endgroup$
– humn
2 hours ago
$begingroup$
Apology for the lack of more specific tags: They would give away the solution.
$endgroup$
– humn
2 hours ago
3
3
$begingroup$
Hurray, a humn puzzle! It's been a while.
$endgroup$
– Rand al'Thor
1 hour ago
$begingroup$
Hurray, a humn puzzle! It's been a while.
$endgroup$
– Rand al'Thor
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The simple relation (for distinct $x$ and $y$)
$x=yiff x,y>10text and x+ytext or x-ytext is a multiple of 20$
tells us that the equivalence class for $20$ is
$20=40=60=80$.
answered 1 hour ago
noednenoedne
7,52212159
$endgroup$
1
$begingroup$
I thought of this too, but it's broken by the fact that 10 and 30 aren't equivalent.
$endgroup$
– Rand al'Thor
1 hour ago
$begingroup$
@Randal'Thor But 10 is not greater than 10.
$endgroup$
– noedne
1 hour ago
$begingroup$
Ah, sorry, my mistake. So there would be one final equivalence class with just four members? My only quibble then is the OP says "not purely mathematical".
$endgroup$
– Rand al'Thor
1 hour ago
$begingroup$
@Randal'Thor You are right, I have removed 100. Any you're right, this is "purely mathematical," but I imagine a non-mathematical answer would be more complicated.
$endgroup$
– noedne
1 hour ago
1
$begingroup$
@noedne, your solution took the bait. A simpler solution is out there.
$endgroup$
– humn
1 hour ago
add a comment |
$begingroup$
20 would be:
20 = 20 = 40 = 40 = 60 = 60 = 80 = 80 = 100
Explanation:
The rule (vertically) is: Line 1 + 1, then Line 2 - 1, and so on.
answered 2 hours ago
XilpexXilpex
235110
New contributor
Xilpex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Thank you for taking the bait, Xilpex. Not quite the solution, though. For instance, it doesn't explain the entry for 10.
$endgroup$
– humn
2 hours ago
$begingroup$
@humn Ok. I'll see if there is any other answer... :D
$endgroup$
– Xilpex
2 hours ago
1
$begingroup$
Plus there is no $100$.
$endgroup$
– Arnaud Mortier
2 hours ago
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
$begingroup$
The simple relation (for distinct $x$ and $y$)
$x=yiff x,y>10text and x+ytext or x-ytext is a multiple of 20$
tells us that the equivalence class for $20$ is
$20=40=60=80$.
answered 1 hour ago
noednenoedne
7,52212159
$endgroup$
1
$begingroup$
I thought of this too, but it's broken by the fact that 10 and 30 aren't equivalent.
$endgroup$
– Rand al'Thor
1 hour ago
$begingroup$
@Randal'Thor But 10 is not greater than 10.
$endgroup$
– noedne
1 hour ago
$begingroup$
Ah, sorry, my mistake. So there would be one final equivalence class with just four members? My only quibble then is the OP says "not purely mathematical".
$endgroup$
– Rand al'Thor
1 hour ago
$begingroup$
@Randal'Thor You are right, I have removed 100. Any you're right, this is "purely mathematical," but I imagine a non-mathematical answer would be more complicated.
$endgroup$
– noedne
1 hour ago
1
$begingroup$
@noedne, your solution took the bait. A simpler solution is out there.
$endgroup$
– humn
1 hour ago
add a comment |
$begingroup$
The simple relation (for distinct $x$ and $y$)
$x=yiff x,y>10text and x+ytext or x-ytext is a multiple of 20$
tells us that the equivalence class for $20$ is
$20=40=60=80$.
answered 1 hour ago
noednenoedne
7,52212159
$endgroup$
1
$begingroup$
I thought of this too, but it's broken by the fact that 10 and 30 aren't equivalent.
$endgroup$
– Rand al'Thor
1 hour ago
$begingroup$
@Randal'Thor But 10 is not greater than 10.
$endgroup$
– noedne
1 hour ago
$begingroup$
Ah, sorry, my mistake. So there would be one final equivalence class with just four members? My only quibble then is the OP says "not purely mathematical".
$endgroup$
– Rand al'Thor
1 hour ago
$begingroup$
@Randal'Thor You are right, I have removed 100. Any you're right, this is "purely mathematical," but I imagine a non-mathematical answer would be more complicated.
$endgroup$
– noedne
1 hour ago
1
$begingroup$
@noedne, your solution took the bait. A simpler solution is out there.
$endgroup$
– humn
1 hour ago
add a comment |
$begingroup$
The simple relation (for distinct $x$ and $y$)
$x=yiff x,y>10text and x+ytext or x-ytext is a multiple of 20$
tells us that the equivalence class for $20$ is
$20=40=60=80$.
answered 1 hour ago
noednenoedne
7,52212159
$endgroup$
The simple relation (for distinct $x$ and $y$)
$x=yiff x,y>10text and x+ytext or x-ytext is a multiple of 20$
tells us that the equivalence class for $20$ is
$20=40=60=80$.
answered 1 hour ago
noednenoedne
7,52212159
edited 1 hour ago
answered 1 hour ago
noednenoedne
7,52212159
answered 1 hour ago
noednenoedne
7,52212159
answered 1 hour ago
noednenoedne
7,52212159
7,52212159
1
$begingroup$
I thought of this too, but it's broken by the fact that 10 and 30 aren't equivalent.
$endgroup$
– Rand al'Thor
1 hour ago
$begingroup$
@Randal'Thor But 10 is not greater than 10.
$endgroup$
– noedne
1 hour ago
$begingroup$
Ah, sorry, my mistake. So there would be one final equivalence class with just four members? My only quibble then is the OP says "not purely mathematical".
$endgroup$
– Rand al'Thor
1 hour ago
$begingroup$
@Randal'Thor You are right, I have removed 100. Any you're right, this is "purely mathematical," but I imagine a non-mathematical answer would be more complicated.
$endgroup$
– noedne
1 hour ago
1
$begingroup$
@noedne, your solution took the bait. A simpler solution is out there.
$endgroup$
– humn
1 hour ago
add a comment |
1
$begingroup$
I thought of this too, but it's broken by the fact that 10 and 30 aren't equivalent.
$endgroup$
– Rand al'Thor
1 hour ago
$begingroup$
@Randal'Thor But 10 is not greater than 10.
$endgroup$
– noedne
1 hour ago
$begingroup$
Ah, sorry, my mistake. So there would be one final equivalence class with just four members? My only quibble then is the OP says "not purely mathematical".
$endgroup$
– Rand al'Thor
1 hour ago
$begingroup$
@Randal'Thor You are right, I have removed 100. Any you're right, this is "purely mathematical," but I imagine a non-mathematical answer would be more complicated.
$endgroup$
– noedne
1 hour ago
1
$begingroup$
@noedne, your solution took the bait. A simpler solution is out there.
$endgroup$
– humn
1 hour ago
1
1
$begingroup$
I thought of this too, but it's broken by the fact that 10 and 30 aren't equivalent.
$endgroup$
– Rand al'Thor
1 hour ago
$begingroup$
I thought of this too, but it's broken by the fact that 10 and 30 aren't equivalent.
$endgroup$
– Rand al'Thor
1 hour ago
$begingroup$
@Randal'Thor But 10 is not greater than 10.
$endgroup$
– noedne
1 hour ago
$begingroup$
@Randal'Thor But 10 is not greater than 10.
$endgroup$
– noedne
1 hour ago
$begingroup$
Ah, sorry, my mistake. So there would be one final equivalence class with just four members? My only quibble then is the OP says "not purely mathematical".
$endgroup$
– Rand al'Thor
1 hour ago
$begingroup$
Ah, sorry, my mistake. So there would be one final equivalence class with just four members? My only quibble then is the OP says "not purely mathematical".
$endgroup$
– Rand al'Thor
1 hour ago
$begingroup$
@Randal'Thor You are right, I have removed 100. Any you're right, this is "purely mathematical," but I imagine a non-mathematical answer would be more complicated.
$endgroup$
– noedne
1 hour ago
$begingroup$
@Randal'Thor You are right, I have removed 100. Any you're right, this is "purely mathematical," but I imagine a non-mathematical answer would be more complicated.
$endgroup$
– noedne
1 hour ago
1
1
$begingroup$
@noedne, your solution took the bait. A simpler solution is out there.
$endgroup$
– humn
1 hour ago
$begingroup$
@noedne, your solution took the bait. A simpler solution is out there.
$endgroup$
– humn
1 hour ago
add a comment |
$begingroup$
20 would be:
20 = 20 = 40 = 40 = 60 = 60 = 80 = 80 = 100
Explanation:
The rule (vertically) is: Line 1 + 1, then Line 2 - 1, and so on.
answered 2 hours ago
XilpexXilpex
235110
New contributor
Xilpex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Thank you for taking the bait, Xilpex. Not quite the solution, though. For instance, it doesn't explain the entry for 10.
$endgroup$
– humn
2 hours ago
$begingroup$
@humn Ok. I'll see if there is any other answer... :D
$endgroup$
– Xilpex
2 hours ago
1
$begingroup$
Plus there is no $100$.
$endgroup$
– Arnaud Mortier
2 hours ago
add a comment |
$begingroup$
20 would be:
20 = 20 = 40 = 40 = 60 = 60 = 80 = 80 = 100
Explanation:
The rule (vertically) is: Line 1 + 1, then Line 2 - 1, and so on.
answered 2 hours ago
XilpexXilpex
235110
New contributor
Xilpex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Thank you for taking the bait, Xilpex. Not quite the solution, though. For instance, it doesn't explain the entry for 10.
$endgroup$
– humn
2 hours ago
$begingroup$
@humn Ok. I'll see if there is any other answer... :D
$endgroup$
– Xilpex
2 hours ago
1
$begingroup$
Plus there is no $100$.
$endgroup$
– Arnaud Mortier
2 hours ago
add a comment |
$begingroup$
20 would be:
20 = 20 = 40 = 40 = 60 = 60 = 80 = 80 = 100
Explanation:
The rule (vertically) is: Line 1 + 1, then Line 2 - 1, and so on.
answered 2 hours ago
XilpexXilpex
235110
New contributor
Xilpex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
20 would be:
20 = 20 = 40 = 40 = 60 = 60 = 80 = 80 = 100
Explanation:
The rule (vertically) is: Line 1 + 1, then Line 2 - 1, and so on.
answered 2 hours ago
XilpexXilpex
235110
New contributor
Xilpex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
XilpexXilpex
235110
New contributor
Xilpex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
XilpexXilpex
235110
answered 2 hours ago
XilpexXilpex
235110
235110
New contributor
Xilpex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Xilpex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Xilpex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Thank you for taking the bait, Xilpex. Not quite the solution, though. For instance, it doesn't explain the entry for 10.
$endgroup$
– humn
2 hours ago
$begingroup$
@humn Ok. I'll see if there is any other answer... :D
$endgroup$
– Xilpex
2 hours ago
1
$begingroup$
Plus there is no $100$.
$endgroup$
– Arnaud Mortier
2 hours ago
add a comment |
1
$begingroup$
Thank you for taking the bait, Xilpex. Not quite the solution, though. For instance, it doesn't explain the entry for 10.
$endgroup$
– humn
2 hours ago
$begingroup$
@humn Ok. I'll see if there is any other answer... :D
$endgroup$
– Xilpex
2 hours ago
1
$begingroup$
Plus there is no $100$.
$endgroup$
– Arnaud Mortier
2 hours ago
1
1
$begingroup$
Thank you for taking the bait, Xilpex. Not quite the solution, though. For instance, it doesn't explain the entry for 10.
$endgroup$
– humn
2 hours ago
$begingroup$
Thank you for taking the bait, Xilpex. Not quite the solution, though. For instance, it doesn't explain the entry for 10.
$endgroup$
– humn
2 hours ago
$begingroup$
@humn Ok. I'll see if there is any other answer... :D
$endgroup$
– Xilpex
2 hours ago
$begingroup$
@humn Ok. I'll see if there is any other answer... :D
$endgroup$
– Xilpex
2 hours ago
1
1
$begingroup$
Plus there is no $100$.
$endgroup$
– Arnaud Mortier
2 hours ago
$begingroup$
Plus there is no $100$.
$endgroup$
– Arnaud Mortier
2 hours ago
add a comment |
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1
$begingroup$
Apology for the lack of more specific tags: They would give away the solution.
$endgroup$
– humn
2 hours ago
3
$begingroup$
Hurray, a humn puzzle! It's been a while.
$endgroup$
– Rand al'Thor
1 hour ago