Mixing
Solving the sequence $3/6, 1/2,3/4/7/6,9/14,3/4/4/5/8/7/7/6,245/256,x,x$
-1
$begingroup$
I need help in solving this numerical sequence. It's a pretty hard one.
Commas divide each part of the sequence. The target is to find the value of the two "$x$"s and write them like a fraction/division.
$$mathbf{3/6}, ;; 1/2, ;; mathbf{3/4/7/6}, ;; 9/14, ;; mathbf{3/4/4/5/8/7/7/6}, ;; 245/256, ;; mathbf{x}, ;; x$$
Problem's owner gave me his permission to repost.
sequences-and-series puzzle
edited Jan 15 at 23:16
Blue
48.5k870154
asked Jan 15 at 22:10
alnesialnesi
14
$endgroup$
|
show 4 more comments
-1
$begingroup$
I need help in solving this numerical sequence. It's a pretty hard one.
Commas divide each part of the sequence. The target is to find the value of the two "$x$"s and write them like a fraction/division.
$$mathbf{3/6}, ;; 1/2, ;; mathbf{3/4/7/6}, ;; 9/14, ;; mathbf{3/4/4/5/8/7/7/6}, ;; 245/256, ;; mathbf{x}, ;; x$$
Problem's owner gave me his permission to repost.
sequences-and-series puzzle
edited Jan 15 at 23:16
Blue
48.5k870154
asked Jan 15 at 22:10
alnesialnesi
14
$endgroup$
3
$begingroup$
you've posted a number of similar problems over the past hour...none of them show any of your own effort and all seem more in the nature of puzzle problems. Perhaps this is not the right site for this sort of question.
$endgroup$
– lulu
Jan 15 at 22:12
$begingroup$
Notation is confusing. What does / stand for?
$endgroup$
– herb steinberg
Jan 15 at 22:14
$begingroup$
Thank you for answering,you are right it's a puzzle but a math puzzle,i've tried to solve those questions in the past days even if i hadn't given enough proof so please be kind ;) @lulu
$endgroup$
– alnesi
Jan 15 at 22:14
$begingroup$
@herbsteinberg those are division where first number is the numerator and the other is the denominator, i tried to rewrite to make it more comprensible, you basically should find the numbers that should be put instead of the x,each comma divides a part of the sequence
$endgroup$
– alnesi
Jan 15 at 22:18
$begingroup$
@herbsteinberg Actually I quite like this - the (deliberately) confusing notation is part of the puzzle. Probably better suited to puzzling.stackexchange though.
$endgroup$
– David
Jan 15 at 22:37
|
show 4 more comments
-1
-1
-1
$begingroup$
I need help in solving this numerical sequence. It's a pretty hard one.
Commas divide each part of the sequence. The target is to find the value of the two "$x$"s and write them like a fraction/division.
$$mathbf{3/6}, ;; 1/2, ;; mathbf{3/4/7/6}, ;; 9/14, ;; mathbf{3/4/4/5/8/7/7/6}, ;; 245/256, ;; mathbf{x}, ;; x$$
Problem's owner gave me his permission to repost.
sequences-and-series puzzle
edited Jan 15 at 23:16
Blue
48.5k870154
asked Jan 15 at 22:10
alnesialnesi
14
$endgroup$
I need help in solving this numerical sequence. It's a pretty hard one.
Commas divide each part of the sequence. The target is to find the value of the two "$x$"s and write them like a fraction/division.
$$mathbf{3/6}, ;; 1/2, ;; mathbf{3/4/7/6}, ;; 9/14, ;; mathbf{3/4/4/5/8/7/7/6}, ;; 245/256, ;; mathbf{x}, ;; x$$
Problem's owner gave me his permission to repost.
sequences-and-series puzzle
sequences-and-series puzzle
edited Jan 15 at 23:16
Blue
48.5k870154
asked Jan 15 at 22:10
alnesialnesi
14
edited Jan 15 at 23:16
Blue
48.5k870154
asked Jan 15 at 22:10
alnesialnesi
14
edited Jan 15 at 23:16
Blue
48.5k870154
edited Jan 15 at 23:16
Blue
48.5k870154
edited Jan 15 at 23:16
Blue
48.5k870154
48.5k870154
asked Jan 15 at 22:10
alnesialnesi
14
asked Jan 15 at 22:10
alnesialnesi
14
asked Jan 15 at 22:10
alnesialnesi
14
14
3
$begingroup$
you've posted a number of similar problems over the past hour...none of them show any of your own effort and all seem more in the nature of puzzle problems. Perhaps this is not the right site for this sort of question.
$endgroup$
– lulu
Jan 15 at 22:12
$begingroup$
Notation is confusing. What does / stand for?
$endgroup$
– herb steinberg
Jan 15 at 22:14
$begingroup$
Thank you for answering,you are right it's a puzzle but a math puzzle,i've tried to solve those questions in the past days even if i hadn't given enough proof so please be kind ;) @lulu
$endgroup$
– alnesi
Jan 15 at 22:14
$begingroup$
@herbsteinberg those are division where first number is the numerator and the other is the denominator, i tried to rewrite to make it more comprensible, you basically should find the numbers that should be put instead of the x,each comma divides a part of the sequence
$endgroup$
– alnesi
Jan 15 at 22:18
$begingroup$
@herbsteinberg Actually I quite like this - the (deliberately) confusing notation is part of the puzzle. Probably better suited to puzzling.stackexchange though.
$endgroup$
– David
Jan 15 at 22:37
|
show 4 more comments
3
$begingroup$
you've posted a number of similar problems over the past hour...none of them show any of your own effort and all seem more in the nature of puzzle problems. Perhaps this is not the right site for this sort of question.
$endgroup$
– lulu
Jan 15 at 22:12
$begingroup$
Notation is confusing. What does / stand for?
$endgroup$
– herb steinberg
Jan 15 at 22:14
$begingroup$
Thank you for answering,you are right it's a puzzle but a math puzzle,i've tried to solve those questions in the past days even if i hadn't given enough proof so please be kind ;) @lulu
$endgroup$
– alnesi
Jan 15 at 22:14
$begingroup$
@herbsteinberg those are division where first number is the numerator and the other is the denominator, i tried to rewrite to make it more comprensible, you basically should find the numbers that should be put instead of the x,each comma divides a part of the sequence
$endgroup$
– alnesi
Jan 15 at 22:18
$begingroup$
@herbsteinberg Actually I quite like this - the (deliberately) confusing notation is part of the puzzle. Probably better suited to puzzling.stackexchange though.
$endgroup$
– David
Jan 15 at 22:37
3
3
$begingroup$
you've posted a number of similar problems over the past hour...none of them show any of your own effort and all seem more in the nature of puzzle problems. Perhaps this is not the right site for this sort of question.
$endgroup$
– lulu
Jan 15 at 22:12
$begingroup$
you've posted a number of similar problems over the past hour...none of them show any of your own effort and all seem more in the nature of puzzle problems. Perhaps this is not the right site for this sort of question.
$endgroup$
– lulu
Jan 15 at 22:12
$begingroup$
Notation is confusing. What does / stand for?
$endgroup$
– herb steinberg
Jan 15 at 22:14
$begingroup$
Notation is confusing. What does / stand for?
$endgroup$
– herb steinberg
Jan 15 at 22:14
$begingroup$
Thank you for answering,you are right it's a puzzle but a math puzzle,i've tried to solve those questions in the past days even if i hadn't given enough proof so please be kind ;) @lulu
$endgroup$
– alnesi
Jan 15 at 22:14
$begingroup$
Thank you for answering,you are right it's a puzzle but a math puzzle,i've tried to solve those questions in the past days even if i hadn't given enough proof so please be kind ;) @lulu
$endgroup$
– alnesi
Jan 15 at 22:14
$begingroup$
@herbsteinberg those are division where first number is the numerator and the other is the denominator, i tried to rewrite to make it more comprensible, you basically should find the numbers that should be put instead of the x,each comma divides a part of the sequence
$endgroup$
– alnesi
Jan 15 at 22:18
$begingroup$
@herbsteinberg those are division where first number is the numerator and the other is the denominator, i tried to rewrite to make it more comprensible, you basically should find the numbers that should be put instead of the x,each comma divides a part of the sequence
$endgroup$
– alnesi
Jan 15 at 22:18
$begingroup$
@herbsteinberg Actually I quite like this - the (deliberately) confusing notation is part of the puzzle. Probably better suited to puzzling.stackexchange though.
$endgroup$
– David
Jan 15 at 22:37
$begingroup$
@herbsteinberg Actually I quite like this - the (deliberately) confusing notation is part of the puzzle. Probably better suited to puzzling.stackexchange though.
$endgroup$
– David
Jan 15 at 22:37
|
show 4 more comments
1 Answer
1
active
oldest
votes
1
$begingroup$
First one: recursion:
$$3/4/4/5/color{red}{4/5/5/6/9/8/8/7/}8/7/7/6$$
Second one: insert brackets:
$$(((3/4)/(4/5))/((4/5)/(5/6)))/(((9/8)/(8/7))/((8/7)/(7/6))) = 1000/1029$$
answered Jan 15 at 22:33
DavidDavid
69k667130
$endgroup$
$begingroup$
I appreciate your effort but i think the solution should be different for one reason,there are 2 X that means that there should be 2 fractions as result,example: 3/6, 1/2, 3/4/7/6, 9/14, 3/4/4/5/8/7/7/6 , 245/256, 10/10, 11/11, I'm just using random numbers to explain what i meant don't take that as the solution. Thank you for you attention @david
$endgroup$
– alnesi
Jan 15 at 22:48
$begingroup$
Where I said "First", that's the first one, where I said "Second" is the second one. Sorry if that was confusing.
$endgroup$
– David
Jan 15 at 23:06
$begingroup$
thank you, i don't understaand where 245/256 went ,did it just disappear? sorry if i'm annoying but want to understand how u reached that result
$endgroup$
– alnesi
Jan 16 at 10:57
$begingroup$
$245/256$ is still there.
$endgroup$
– David
Jan 16 at 23:03
$begingroup$
Maybe i'm stupid but i don't understand your solution,could you provide a longer answer there please? IN this sequence 3/6,1/2,3/4/7/6,9/14,3/4/4/5/8/7/7/6,245/256,x,x there are 2 x that means we should get 2 result ,but you wrote 1000/1029 and that's only one solution,please explain me that i'm a newbie
$endgroup$
– alnesi
Jan 20 at 13:49
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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1
$begingroup$
First one: recursion:
$$3/4/4/5/color{red}{4/5/5/6/9/8/8/7/}8/7/7/6$$
Second one: insert brackets:
$$(((3/4)/(4/5))/((4/5)/(5/6)))/(((9/8)/(8/7))/((8/7)/(7/6))) = 1000/1029$$
answered Jan 15 at 22:33
DavidDavid
69k667130
$endgroup$
$begingroup$
I appreciate your effort but i think the solution should be different for one reason,there are 2 X that means that there should be 2 fractions as result,example: 3/6, 1/2, 3/4/7/6, 9/14, 3/4/4/5/8/7/7/6 , 245/256, 10/10, 11/11, I'm just using random numbers to explain what i meant don't take that as the solution. Thank you for you attention @david
$endgroup$
– alnesi
Jan 15 at 22:48
$begingroup$
Where I said "First", that's the first one, where I said "Second" is the second one. Sorry if that was confusing.
$endgroup$
– David
Jan 15 at 23:06
$begingroup$
thank you, i don't understaand where 245/256 went ,did it just disappear? sorry if i'm annoying but want to understand how u reached that result
$endgroup$
– alnesi
Jan 16 at 10:57
$begingroup$
$245/256$ is still there.
$endgroup$
– David
Jan 16 at 23:03
$begingroup$
Maybe i'm stupid but i don't understand your solution,could you provide a longer answer there please? IN this sequence 3/6,1/2,3/4/7/6,9/14,3/4/4/5/8/7/7/6,245/256,x,x there are 2 x that means we should get 2 result ,but you wrote 1000/1029 and that's only one solution,please explain me that i'm a newbie
$endgroup$
– alnesi
Jan 20 at 13:49
add a comment |
1
$begingroup$
First one: recursion:
$$3/4/4/5/color{red}{4/5/5/6/9/8/8/7/}8/7/7/6$$
Second one: insert brackets:
$$(((3/4)/(4/5))/((4/5)/(5/6)))/(((9/8)/(8/7))/((8/7)/(7/6))) = 1000/1029$$
answered Jan 15 at 22:33
DavidDavid
69k667130
$endgroup$
$begingroup$
I appreciate your effort but i think the solution should be different for one reason,there are 2 X that means that there should be 2 fractions as result,example: 3/6, 1/2, 3/4/7/6, 9/14, 3/4/4/5/8/7/7/6 , 245/256, 10/10, 11/11, I'm just using random numbers to explain what i meant don't take that as the solution. Thank you for you attention @david
$endgroup$
– alnesi
Jan 15 at 22:48
$begingroup$
Where I said "First", that's the first one, where I said "Second" is the second one. Sorry if that was confusing.
$endgroup$
– David
Jan 15 at 23:06
$begingroup$
thank you, i don't understaand where 245/256 went ,did it just disappear? sorry if i'm annoying but want to understand how u reached that result
$endgroup$
– alnesi
Jan 16 at 10:57
$begingroup$
$245/256$ is still there.
$endgroup$
– David
Jan 16 at 23:03
$begingroup$
Maybe i'm stupid but i don't understand your solution,could you provide a longer answer there please? IN this sequence 3/6,1/2,3/4/7/6,9/14,3/4/4/5/8/7/7/6,245/256,x,x there are 2 x that means we should get 2 result ,but you wrote 1000/1029 and that's only one solution,please explain me that i'm a newbie
$endgroup$
– alnesi
Jan 20 at 13:49
add a comment |
1
1
1
$begingroup$
First one: recursion:
$$3/4/4/5/color{red}{4/5/5/6/9/8/8/7/}8/7/7/6$$
Second one: insert brackets:
$$(((3/4)/(4/5))/((4/5)/(5/6)))/(((9/8)/(8/7))/((8/7)/(7/6))) = 1000/1029$$
answered Jan 15 at 22:33
DavidDavid
69k667130
$endgroup$
First one: recursion:
$$3/4/4/5/color{red}{4/5/5/6/9/8/8/7/}8/7/7/6$$
Second one: insert brackets:
$$(((3/4)/(4/5))/((4/5)/(5/6)))/(((9/8)/(8/7))/((8/7)/(7/6))) = 1000/1029$$
answered Jan 15 at 22:33
DavidDavid
69k667130
edited Jan 15 at 23:05
answered Jan 15 at 22:33
DavidDavid
69k667130
answered Jan 15 at 22:33
DavidDavid
69k667130
answered Jan 15 at 22:33
DavidDavid
69k667130
69k667130
$begingroup$
I appreciate your effort but i think the solution should be different for one reason,there are 2 X that means that there should be 2 fractions as result,example: 3/6, 1/2, 3/4/7/6, 9/14, 3/4/4/5/8/7/7/6 , 245/256, 10/10, 11/11, I'm just using random numbers to explain what i meant don't take that as the solution. Thank you for you attention @david
$endgroup$
– alnesi
Jan 15 at 22:48
$begingroup$
Where I said "First", that's the first one, where I said "Second" is the second one. Sorry if that was confusing.
$endgroup$
– David
Jan 15 at 23:06
$begingroup$
thank you, i don't understaand where 245/256 went ,did it just disappear? sorry if i'm annoying but want to understand how u reached that result
$endgroup$
– alnesi
Jan 16 at 10:57
$begingroup$
$245/256$ is still there.
$endgroup$
– David
Jan 16 at 23:03
$begingroup$
Maybe i'm stupid but i don't understand your solution,could you provide a longer answer there please? IN this sequence 3/6,1/2,3/4/7/6,9/14,3/4/4/5/8/7/7/6,245/256,x,x there are 2 x that means we should get 2 result ,but you wrote 1000/1029 and that's only one solution,please explain me that i'm a newbie
$endgroup$
– alnesi
Jan 20 at 13:49
add a comment |
$begingroup$
I appreciate your effort but i think the solution should be different for one reason,there are 2 X that means that there should be 2 fractions as result,example: 3/6, 1/2, 3/4/7/6, 9/14, 3/4/4/5/8/7/7/6 , 245/256, 10/10, 11/11, I'm just using random numbers to explain what i meant don't take that as the solution. Thank you for you attention @david
$endgroup$
– alnesi
Jan 15 at 22:48
$begingroup$
Where I said "First", that's the first one, where I said "Second" is the second one. Sorry if that was confusing.
$endgroup$
– David
Jan 15 at 23:06
$begingroup$
thank you, i don't understaand where 245/256 went ,did it just disappear? sorry if i'm annoying but want to understand how u reached that result
$endgroup$
– alnesi
Jan 16 at 10:57
$begingroup$
$245/256$ is still there.
$endgroup$
– David
Jan 16 at 23:03
$begingroup$
Maybe i'm stupid but i don't understand your solution,could you provide a longer answer there please? IN this sequence 3/6,1/2,3/4/7/6,9/14,3/4/4/5/8/7/7/6,245/256,x,x there are 2 x that means we should get 2 result ,but you wrote 1000/1029 and that's only one solution,please explain me that i'm a newbie
$endgroup$
– alnesi
Jan 20 at 13:49
$begingroup$
I appreciate your effort but i think the solution should be different for one reason,there are 2 X that means that there should be 2 fractions as result,example: 3/6, 1/2, 3/4/7/6, 9/14, 3/4/4/5/8/7/7/6 , 245/256, 10/10, 11/11, I'm just using random numbers to explain what i meant don't take that as the solution. Thank you for you attention @david
$endgroup$
– alnesi
Jan 15 at 22:48
$begingroup$
I appreciate your effort but i think the solution should be different for one reason,there are 2 X that means that there should be 2 fractions as result,example: 3/6, 1/2, 3/4/7/6, 9/14, 3/4/4/5/8/7/7/6 , 245/256, 10/10, 11/11, I'm just using random numbers to explain what i meant don't take that as the solution. Thank you for you attention @david
$endgroup$
– alnesi
Jan 15 at 22:48
$begingroup$
Where I said "First", that's the first one, where I said "Second" is the second one. Sorry if that was confusing.
$endgroup$
– David
Jan 15 at 23:06
$begingroup$
Where I said "First", that's the first one, where I said "Second" is the second one. Sorry if that was confusing.
$endgroup$
– David
Jan 15 at 23:06
$begingroup$
thank you, i don't understaand where 245/256 went ,did it just disappear? sorry if i'm annoying but want to understand how u reached that result
$endgroup$
– alnesi
Jan 16 at 10:57
$begingroup$
thank you, i don't understaand where 245/256 went ,did it just disappear? sorry if i'm annoying but want to understand how u reached that result
$endgroup$
– alnesi
Jan 16 at 10:57
$begingroup$
$245/256$ is still there.
$endgroup$
– David
Jan 16 at 23:03
$begingroup$
$245/256$ is still there.
$endgroup$
– David
Jan 16 at 23:03
$begingroup$
Maybe i'm stupid but i don't understand your solution,could you provide a longer answer there please? IN this sequence 3/6,1/2,3/4/7/6,9/14,3/4/4/5/8/7/7/6,245/256,x,x there are 2 x that means we should get 2 result ,but you wrote 1000/1029 and that's only one solution,please explain me that i'm a newbie
$endgroup$
– alnesi
Jan 20 at 13:49
$begingroup$
Maybe i'm stupid but i don't understand your solution,could you provide a longer answer there please? IN this sequence 3/6,1/2,3/4/7/6,9/14,3/4/4/5/8/7/7/6,245/256,x,x there are 2 x that means we should get 2 result ,but you wrote 1000/1029 and that's only one solution,please explain me that i'm a newbie
$endgroup$
– alnesi
Jan 20 at 13:49
add a comment |
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3
$begingroup$
you've posted a number of similar problems over the past hour...none of them show any of your own effort and all seem more in the nature of puzzle problems. Perhaps this is not the right site for this sort of question.
$endgroup$
– lulu
Jan 15 at 22:12
$begingroup$
Notation is confusing. What does / stand for?
$endgroup$
– herb steinberg
Jan 15 at 22:14
$begingroup$
Thank you for answering,you are right it's a puzzle but a math puzzle,i've tried to solve those questions in the past days even if i hadn't given enough proof so please be kind ;) @lulu
$endgroup$
– alnesi
Jan 15 at 22:14
$begingroup$
@herbsteinberg those are division where first number is the numerator and the other is the denominator, i tried to rewrite to make it more comprensible, you basically should find the numbers that should be put instead of the x,each comma divides a part of the sequence
$endgroup$
– alnesi
Jan 15 at 22:18
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@herbsteinberg Actually I quite like this - the (deliberately) confusing notation is part of the puzzle. Probably better suited to puzzling.stackexchange though.
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– David
Jan 15 at 22:37