Unlock my phone! March 2018Which Flight Takes Me Home?A puzzle inspired by “Cryptobiology? Biocryptology?”A puzzle-loving lady's phone numberIs it a bird? Is it a plane? No, it's a comet!Lost my Password! Please help!one riddle that is hiddenMysterious Flash DriveBro's password… must've forgottenUnlock My Phone! January 2018Unlock My Phone! February 2018
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Unlock my phone! March 2018
Which Flight Takes Me Home?A puzzle inspired by “Cryptobiology? Biocryptology?”A puzzle-loving lady's phone numberIs it a bird? Is it a plane? No, it's a comet!Lost my Password! Please help!one riddle that is hiddenMysterious Flash DriveBro's password… must've forgottenUnlock My Phone! January 2018Unlock My Phone! February 2018
$begingroup$
New month, new password, but I've already forgotten! Help!
What is my phone's password this month?
Hint #1:
As per usual, my password is pretty long (maybe around 40 digits?)
Hint #2:
In March I took up an interest in analytical number theory
mathematics knowledge visual
asked 8 hours ago
Tyler22AlexTyler22Alex
184110
New contributor
Tyler22Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
New month, new password, but I've already forgotten! Help!
What is my phone's password this month?
Hint #1:
As per usual, my password is pretty long (maybe around 40 digits?)
Hint #2:
In March I took up an interest in analytical number theory
mathematics knowledge visual
asked 8 hours ago
Tyler22AlexTyler22Alex
184110
New contributor
Tyler22Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Really March? Really 2018?
$endgroup$
– Gareth McCaughan♦
8 hours ago
$begingroup$
If you look at my other puzzles you’ll see I’m starting January of 2018 and going month by month to the present day. Sorry for confusion!
$endgroup$
– Tyler22Alex
7 hours ago
$begingroup$
Ah, OK. Fair enough.
$endgroup$
– Gareth McCaughan♦
6 hours ago
add a comment |
$begingroup$
New month, new password, but I've already forgotten! Help!
What is my phone's password this month?
Hint #1:
As per usual, my password is pretty long (maybe around 40 digits?)
Hint #2:
In March I took up an interest in analytical number theory
mathematics knowledge visual
asked 8 hours ago
Tyler22AlexTyler22Alex
184110
New contributor
Tyler22Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
New month, new password, but I've already forgotten! Help!
What is my phone's password this month?
Hint #1:
As per usual, my password is pretty long (maybe around 40 digits?)
Hint #2:
In March I took up an interest in analytical number theory
mathematics knowledge visual
mathematics knowledge visual
asked 8 hours ago
Tyler22AlexTyler22Alex
184110
New contributor
Tyler22Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
Tyler22AlexTyler22Alex
184110
New contributor
Tyler22Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
Tyler22AlexTyler22Alex
184110
New contributor
Tyler22Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
Tyler22AlexTyler22Alex
184110
asked 8 hours ago
Tyler22AlexTyler22Alex
184110
184110
New contributor
Tyler22Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Tyler22Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Tyler22Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Really March? Really 2018?
$endgroup$
– Gareth McCaughan♦
8 hours ago
$begingroup$
If you look at my other puzzles you’ll see I’m starting January of 2018 and going month by month to the present day. Sorry for confusion!
$endgroup$
– Tyler22Alex
7 hours ago
$begingroup$
Ah, OK. Fair enough.
$endgroup$
– Gareth McCaughan♦
6 hours ago
add a comment |
1
$begingroup$
Really March? Really 2018?
$endgroup$
– Gareth McCaughan♦
8 hours ago
$begingroup$
If you look at my other puzzles you’ll see I’m starting January of 2018 and going month by month to the present day. Sorry for confusion!
$endgroup$
– Tyler22Alex
7 hours ago
$begingroup$
Ah, OK. Fair enough.
$endgroup$
– Gareth McCaughan♦
6 hours ago
1
1
$begingroup$
Really March? Really 2018?
$endgroup$
– Gareth McCaughan♦
8 hours ago
$begingroup$
Really March? Really 2018?
$endgroup$
– Gareth McCaughan♦
8 hours ago
$begingroup$
If you look at my other puzzles you’ll see I’m starting January of 2018 and going month by month to the present day. Sorry for confusion!
$endgroup$
– Tyler22Alex
7 hours ago
$begingroup$
If you look at my other puzzles you’ll see I’m starting January of 2018 and going month by month to the present day. Sorry for confusion!
$endgroup$
– Tyler22Alex
7 hours ago
$begingroup$
Ah, OK. Fair enough.
$endgroup$
– Gareth McCaughan♦
6 hours ago
$begingroup$
Ah, OK. Fair enough.
$endgroup$
– Gareth McCaughan♦
6 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your password is
the first 40 digits, or perhaps decimal places, of the Euler-Mascheroni constant, sometimes denoted $gamma$.
So it will begin
either 5772 or 05772 or 0.5772.
Since
to 41 decimal places the constant's value is 0.57721566490153286060651209008240243104215, the last digit will (if you've used 40 decimal places rather than 40 digits including the leading zero) depend on how you've chosen to round it.
Explanation in case the above doesn't make it clear:
the harmonic series is $1+frac12+frac13+frac14+cdots$ and the sum of its first $n$ terms is $log n+gamma+varepsilon(n)$ where $varepsilon(n)rightarrow0$ as $nrightarrowinfty$. So the limiting difference between the harmonic series and the natural logarithm is $gamma$.
answered 8 hours ago
Gareth McCaughan♦Gareth McCaughan
66.2k3167259
$endgroup$
1
$begingroup$
That was fast! Nice job!
$endgroup$
– Tyler22Alex
7 hours ago
$begingroup$
The second hint made it pretty hard not to get :-).
$endgroup$
– Gareth McCaughan♦
6 hours ago
$begingroup$
(Though I think I'd probably have got it quickly even without.)
$endgroup$
– Gareth McCaughan♦
6 hours ago
add a comment |
Your Answer
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1 Answer
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votes
$begingroup$
Your password is
the first 40 digits, or perhaps decimal places, of the Euler-Mascheroni constant, sometimes denoted $gamma$.
So it will begin
either 5772 or 05772 or 0.5772.
Since
to 41 decimal places the constant's value is 0.57721566490153286060651209008240243104215, the last digit will (if you've used 40 decimal places rather than 40 digits including the leading zero) depend on how you've chosen to round it.
Explanation in case the above doesn't make it clear:
the harmonic series is $1+frac12+frac13+frac14+cdots$ and the sum of its first $n$ terms is $log n+gamma+varepsilon(n)$ where $varepsilon(n)rightarrow0$ as $nrightarrowinfty$. So the limiting difference between the harmonic series and the natural logarithm is $gamma$.
answered 8 hours ago
Gareth McCaughan♦Gareth McCaughan
66.2k3167259
$endgroup$
1
$begingroup$
That was fast! Nice job!
$endgroup$
– Tyler22Alex
7 hours ago
$begingroup$
The second hint made it pretty hard not to get :-).
$endgroup$
– Gareth McCaughan♦
6 hours ago
$begingroup$
(Though I think I'd probably have got it quickly even without.)
$endgroup$
– Gareth McCaughan♦
6 hours ago
add a comment |
$begingroup$
Your password is
the first 40 digits, or perhaps decimal places, of the Euler-Mascheroni constant, sometimes denoted $gamma$.
So it will begin
either 5772 or 05772 or 0.5772.
Since
to 41 decimal places the constant's value is 0.57721566490153286060651209008240243104215, the last digit will (if you've used 40 decimal places rather than 40 digits including the leading zero) depend on how you've chosen to round it.
Explanation in case the above doesn't make it clear:
the harmonic series is $1+frac12+frac13+frac14+cdots$ and the sum of its first $n$ terms is $log n+gamma+varepsilon(n)$ where $varepsilon(n)rightarrow0$ as $nrightarrowinfty$. So the limiting difference between the harmonic series and the natural logarithm is $gamma$.
answered 8 hours ago
Gareth McCaughan♦Gareth McCaughan
66.2k3167259
$endgroup$
1
$begingroup$
That was fast! Nice job!
$endgroup$
– Tyler22Alex
7 hours ago
$begingroup$
The second hint made it pretty hard not to get :-).
$endgroup$
– Gareth McCaughan♦
6 hours ago
$begingroup$
(Though I think I'd probably have got it quickly even without.)
$endgroup$
– Gareth McCaughan♦
6 hours ago
add a comment |
$begingroup$
Your password is
the first 40 digits, or perhaps decimal places, of the Euler-Mascheroni constant, sometimes denoted $gamma$.
So it will begin
either 5772 or 05772 or 0.5772.
Since
to 41 decimal places the constant's value is 0.57721566490153286060651209008240243104215, the last digit will (if you've used 40 decimal places rather than 40 digits including the leading zero) depend on how you've chosen to round it.
Explanation in case the above doesn't make it clear:
the harmonic series is $1+frac12+frac13+frac14+cdots$ and the sum of its first $n$ terms is $log n+gamma+varepsilon(n)$ where $varepsilon(n)rightarrow0$ as $nrightarrowinfty$. So the limiting difference between the harmonic series and the natural logarithm is $gamma$.
answered 8 hours ago
Gareth McCaughan♦Gareth McCaughan
66.2k3167259
$endgroup$
Your password is
the first 40 digits, or perhaps decimal places, of the Euler-Mascheroni constant, sometimes denoted $gamma$.
So it will begin
either 5772 or 05772 or 0.5772.
Since
to 41 decimal places the constant's value is 0.57721566490153286060651209008240243104215, the last digit will (if you've used 40 decimal places rather than 40 digits including the leading zero) depend on how you've chosen to round it.
Explanation in case the above doesn't make it clear:
the harmonic series is $1+frac12+frac13+frac14+cdots$ and the sum of its first $n$ terms is $log n+gamma+varepsilon(n)$ where $varepsilon(n)rightarrow0$ as $nrightarrowinfty$. So the limiting difference between the harmonic series and the natural logarithm is $gamma$.
answered 8 hours ago
Gareth McCaughan♦Gareth McCaughan
66.2k3167259
answered 8 hours ago
Gareth McCaughan♦Gareth McCaughan
66.2k3167259
answered 8 hours ago
Gareth McCaughan♦Gareth McCaughan
66.2k3167259
answered 8 hours ago
Gareth McCaughan♦Gareth McCaughan
66.2k3167259
66.2k3167259
1
$begingroup$
That was fast! Nice job!
$endgroup$
– Tyler22Alex
7 hours ago
$begingroup$
The second hint made it pretty hard not to get :-).
$endgroup$
– Gareth McCaughan♦
6 hours ago
$begingroup$
(Though I think I'd probably have got it quickly even without.)
$endgroup$
– Gareth McCaughan♦
6 hours ago
add a comment |
1
$begingroup$
That was fast! Nice job!
$endgroup$
– Tyler22Alex
7 hours ago
$begingroup$
The second hint made it pretty hard not to get :-).
$endgroup$
– Gareth McCaughan♦
6 hours ago
$begingroup$
(Though I think I'd probably have got it quickly even without.)
$endgroup$
– Gareth McCaughan♦
6 hours ago
1
1
$begingroup$
That was fast! Nice job!
$endgroup$
– Tyler22Alex
7 hours ago
$begingroup$
That was fast! Nice job!
$endgroup$
– Tyler22Alex
7 hours ago
$begingroup$
The second hint made it pretty hard not to get :-).
$endgroup$
– Gareth McCaughan♦
6 hours ago
$begingroup$
The second hint made it pretty hard not to get :-).
$endgroup$
– Gareth McCaughan♦
6 hours ago
$begingroup$
(Though I think I'd probably have got it quickly even without.)
$endgroup$
– Gareth McCaughan♦
6 hours ago
$begingroup$
(Though I think I'd probably have got it quickly even without.)
$endgroup$
– Gareth McCaughan♦
6 hours ago
add a comment |
Tyler22Alex is a new contributor. Be nice, and check out our Code of Conduct.
Tyler22Alex is a new contributor. Be nice, and check out our Code of Conduct.
Tyler22Alex is a new contributor. Be nice, and check out our Code of Conduct.
Tyler22Alex is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
Really March? Really 2018?
$endgroup$
– Gareth McCaughan♦
8 hours ago
$begingroup$
If you look at my other puzzles you’ll see I’m starting January of 2018 and going month by month to the present day. Sorry for confusion!
$endgroup$
– Tyler22Alex
7 hours ago
$begingroup$
Ah, OK. Fair enough.
$endgroup$
– Gareth McCaughan♦
6 hours ago