Unlock My Phone! February 2018
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I forgot my password again! I changed it a few days ago but already forgot!
What is my phone's password this month?
Hint #1:
My passwords do tend to be fairly long, as mentioned in the comments.
Hint #2:
The number itself may not have been on the moon, but maybe something it shares a trait with.
knowledge visual
asked Apr 2 at 17:08
Tyler22AlexTyler22Alex
313211
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Tyler22Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
I forgot my password again! I changed it a few days ago but already forgot!
What is my phone's password this month?
Hint #1:
My passwords do tend to be fairly long, as mentioned in the comments.
Hint #2:
The number itself may not have been on the moon, but maybe something it shares a trait with.
knowledge visual
asked Apr 2 at 17:08
Tyler22AlexTyler22Alex
313211
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
$begingroup$
As a small recommendation, linked with Rubio's answer, if the January 2018 puzzle's hint also applies to this question, it would be better to also place it. It's better to get a repeated hint than a missing yet important hint
$endgroup$
– Belhenix
Apr 2 at 22:30
2
$begingroup$
@Belhenix I added that hint as well as one more. I'll keep this in mind for all future puzzles too, thanks!
$endgroup$
– Tyler22Alex
Apr 3 at 0:32
add a comment |
$begingroup$
I forgot my password again! I changed it a few days ago but already forgot!
What is my phone's password this month?
Hint #1:
My passwords do tend to be fairly long, as mentioned in the comments.
Hint #2:
The number itself may not have been on the moon, but maybe something it shares a trait with.
knowledge visual
asked Apr 2 at 17:08
Tyler22AlexTyler22Alex
313211
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.
$endgroup$
I forgot my password again! I changed it a few days ago but already forgot!
What is my phone's password this month?
Hint #1:
My passwords do tend to be fairly long, as mentioned in the comments.
Hint #2:
The number itself may not have been on the moon, but maybe something it shares a trait with.
knowledge visual
knowledge visual
asked Apr 2 at 17:08
Tyler22AlexTyler22Alex
313211
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 Apr 2 at 17:08
Tyler22AlexTyler22Alex
313211
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.
edited Apr 3 at 0:26
Tyler22Alex
asked Apr 2 at 17:08
Tyler22AlexTyler22Alex
313211
New contributor
Tyler22Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Apr 2 at 17:08
Tyler22AlexTyler22Alex
313211
asked Apr 2 at 17:08
Tyler22AlexTyler22Alex
313211
313211
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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.
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Check out our Code of Conduct.
1
$begingroup$
As a small recommendation, linked with Rubio's answer, if the January 2018 puzzle's hint also applies to this question, it would be better to also place it. It's better to get a repeated hint than a missing yet important hint
$endgroup$
– Belhenix
Apr 2 at 22:30
2
$begingroup$
@Belhenix I added that hint as well as one more. I'll keep this in mind for all future puzzles too, thanks!
$endgroup$
– Tyler22Alex
Apr 3 at 0:32
add a comment |
1
$begingroup$
As a small recommendation, linked with Rubio's answer, if the January 2018 puzzle's hint also applies to this question, it would be better to also place it. It's better to get a repeated hint than a missing yet important hint
$endgroup$
– Belhenix
Apr 2 at 22:30
2
$begingroup$
@Belhenix I added that hint as well as one more. I'll keep this in mind for all future puzzles too, thanks!
$endgroup$
– Tyler22Alex
Apr 3 at 0:32
1
1
$begingroup$
As a small recommendation, linked with Rubio's answer, if the January 2018 puzzle's hint also applies to this question, it would be better to also place it. It's better to get a repeated hint than a missing yet important hint
$endgroup$
– Belhenix
Apr 2 at 22:30
$begingroup$
As a small recommendation, linked with Rubio's answer, if the January 2018 puzzle's hint also applies to this question, it would be better to also place it. It's better to get a repeated hint than a missing yet important hint
$endgroup$
– Belhenix
Apr 2 at 22:30
2
2
$begingroup$
@Belhenix I added that hint as well as one more. I'll keep this in mind for all future puzzles too, thanks!
$endgroup$
– Tyler22Alex
Apr 3 at 0:32
$begingroup$
@Belhenix I added that hint as well as one more. I'll keep this in mind for all future puzzles too, thanks!
$endgroup$
– Tyler22Alex
Apr 3 at 0:32
add a comment |
8 Answers
8
active
oldest
votes
$begingroup$
Your lock code is
115132219018763992565095597973971522401
The reason
Neil Armstrong was the first man on moon. Armstrong number is an n-digit number in which sum of n-th power of each digit is equal to the number itself.
E.g. - 371 = 3^3 + 7^3 + 1^3. (3 digit number so power of 3).
Using this rationale, we find that
115132219018763992565095597973971522401 is the largest known armstrong number as per this and has 39 digits.
No wonder you forgot it!
Bonus reading - wikipedia
answered 2 days ago
RedBaronRedBaron
33626
New contributor
RedBaron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
5
$begingroup$
Nice job! You got it! Side note: I actually used this (and all other password puzzles I post) as my password and I don't think I've forgotten one yet :) I like numbers!
$endgroup$
– Tyler22Alex
2 days ago
add a comment |
$begingroup$
I believe the answer is
1000
I think that we are supposed to take this literally as
"first of the word moon, which is M (1000 in Roman)"
answered Apr 2 at 19:49
AProughAPrough
5,9061245
$endgroup$
1
$begingroup$
A tag of the puzzle is knowledge. (ROT13) Oryvrir vg be abg, lbh pna gnxr bhg yrggref sebz "Ebzna Xabjyrqtr" arprffnel gb znxr gur jbeq "zbba". Creuncf lbh ner ba gur evtug genpx, ubjrire gur ahzore zvtug or n ovg fznyy nppbeqvat gb Uvag 1. $(+1)$ though!
$endgroup$
– user477343
Apr 3 at 0:36
1
$begingroup$
Very interesting answer, but not quite! Maybe take a look at the hints I added?
$endgroup$
– Tyler22Alex
Apr 3 at 0:44
$begingroup$
Yeah, answer was from before the hints. Nice puzzle though.
$endgroup$
– APrough
2 days ago
add a comment |
$begingroup$
Okay, so first of all, I checked the image and contrast to see if there was something hidden there. I couldn't find anything. Next, I followed the literal interpratation of the text, that is:
The first man to walk on the moon was Neil Armstrong. I tried looking for any items that he wore that would have a serial number or something like that. According to this site, https://www.fratellowatches.com/speedy-tuesday-speedmaster-professional-105-012-with-serial-number-2400xxxx/, the Speedwatches were the only ones used on the moon.
Neil Armstrong's serial number on his watch (again, according to the site) is 24002981, which I believe is your password.
answered Apr 2 at 17:25
Joe-You-KnowJoe-You-Know
6,59821073
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7
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As a number, you don't need people onboard to be first on the moon.
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– Arnaud Mortier
Apr 2 at 18:20
1
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On the right track with Armstrong, but looking in the wrong direction for the number. Maybe what you've got so far and my second hint will get you closer!
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– Tyler22Alex
2 days ago
add a comment |
$begingroup$
Perhaps the password is
1959
Explanation:
Luna 2 was the first spacecraft to make contact with the moon.
According to Wikipedia, Luna 2 was engraved with "1959 январь" and "СОЮЗ СОВЕТСКИХ СОЦИАЛИСТИЧЕСКИХ РЕСПУБЛИК"1, making "1959" the first number on the moon.
Furthermore, phone passcodes are typically 4 digits long, so using a year as a passcode makes sense.
answered Apr 2 at 21:21
user1812user1812
912
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1
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Pursuant to Hint 1 however, this does not quite match, thus making it unlikely to be the answer; nevertheless, I like it! $(+1)$ :P
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– user477343
Apr 3 at 0:28
1
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I think this was posted before I added the first hint, so I apologize for that, but good thinking nonetheless!
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– Tyler22Alex
2 days ago
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@Tyler22Alex oh, it was? My bad. (Also, is it nevertheless or nonetheless...? Or are they interchangeable?)
$endgroup$
– user477343
2 days ago
add a comment |
$begingroup$
Perhaps the password is
11 or eleven or some variation of it
because
Apollo 11 was the first manned mission to the moon, and it had the “largest number” of anything that was first to the moon, greater that Luna 2, which was the first successful unmanned spacecraft to land.
answered Apr 2 at 19:07
PiIsNot3PiIsNot3
1,61222
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1
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I see the number in Apollo! $(+1)$
$endgroup$
– user477343
Apr 3 at 0:37
1
$begingroup$
Interesting idea! However I think you posted this before I added the first hint, so I apologize for that.
$endgroup$
– Tyler22Alex
2 days ago
add a comment |
$begingroup$
There are a lot of ways to find numbers that are "first on the moon", so this question may need some more specificity to rule out answers; right now there doesn't seem to be any criteria beyond "the largest" number an answerer might justify somehow, which is probably too broad.
It's possible the password is
201901030226
We know that
OP uses passwords that are usually pretty long, as we learned in Unlock My Phone! January 2018.
We're looking for the largest "first on the moon" number.
China recently made the first landing ever on the far side of the moon - the robotic
Chang'e 4 mission touched down in the Von Kármán Crater at 10:26am Jan. 3, Beijing time.
This is clearly a "first" on the moon, both for China (first moon landing at all) and for humanity (first landing on the dark side of the moon).
If we take the date and time, in GMT, of this moon landing, we get 201901030226. Seconds were not given, so this is the most specific timedate stamp we can get.
Since this is the most recent moon landing, if we were to consider any other "first" - by nation, by mission type, whatever - and take their corresponding YYYYMMDDHMM number, this one would be the largest.
answered Apr 2 at 22:14
Rubio♦Rubio
30.5k567188
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1
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Hopefully my hint helped a bit, and I'll add more hints as time goes by if people still aren't getting it.
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– Tyler22Alex
2 days ago
add a comment |
$begingroup$
I think it's:
23.9 billion
Explanation:
According to Wikipedia:
In January 1969, NASA prepared an itemized estimate of the run-out cost of the Apollo program. The total came to $23.9 billion
answered Apr 2 at 21:05
tuvokkituvokki
1312
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1
$begingroup$
Good thinking but not quite! Maybe look at the hints I added. Good luck!
$endgroup$
– Tyler22Alex
2 days ago
add a comment |
$begingroup$
The answer is
9.5 Because it's the boot size Neil Armstrong had when first setting foot on the moon.
Thus it was "the largest number to be first on the moon"
answered 2 days ago
PeterPeter
2006
$endgroup$
add a comment |
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8 Answers
8
active
oldest
votes
8 Answers
8
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your lock code is
115132219018763992565095597973971522401
The reason
Neil Armstrong was the first man on moon. Armstrong number is an n-digit number in which sum of n-th power of each digit is equal to the number itself.
E.g. - 371 = 3^3 + 7^3 + 1^3. (3 digit number so power of 3).
Using this rationale, we find that
115132219018763992565095597973971522401 is the largest known armstrong number as per this and has 39 digits.
No wonder you forgot it!
Bonus reading - wikipedia
answered 2 days ago
RedBaronRedBaron
33626
New contributor
RedBaron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
5
$begingroup$
Nice job! You got it! Side note: I actually used this (and all other password puzzles I post) as my password and I don't think I've forgotten one yet :) I like numbers!
$endgroup$
– Tyler22Alex
2 days ago
add a comment |
$begingroup$
Your lock code is
115132219018763992565095597973971522401
The reason
Neil Armstrong was the first man on moon. Armstrong number is an n-digit number in which sum of n-th power of each digit is equal to the number itself.
E.g. - 371 = 3^3 + 7^3 + 1^3. (3 digit number so power of 3).
Using this rationale, we find that
115132219018763992565095597973971522401 is the largest known armstrong number as per this and has 39 digits.
No wonder you forgot it!
Bonus reading - wikipedia
answered 2 days ago
RedBaronRedBaron
33626
New contributor
RedBaron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
5
$begingroup$
Nice job! You got it! Side note: I actually used this (and all other password puzzles I post) as my password and I don't think I've forgotten one yet :) I like numbers!
$endgroup$
– Tyler22Alex
2 days ago
add a comment |
$begingroup$
Your lock code is
115132219018763992565095597973971522401
The reason
Neil Armstrong was the first man on moon. Armstrong number is an n-digit number in which sum of n-th power of each digit is equal to the number itself.
E.g. - 371 = 3^3 + 7^3 + 1^3. (3 digit number so power of 3).
Using this rationale, we find that
115132219018763992565095597973971522401 is the largest known armstrong number as per this and has 39 digits.
No wonder you forgot it!
Bonus reading - wikipedia
answered 2 days ago
RedBaronRedBaron
33626
New contributor
RedBaron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Your lock code is
115132219018763992565095597973971522401
The reason
Neil Armstrong was the first man on moon. Armstrong number is an n-digit number in which sum of n-th power of each digit is equal to the number itself.
E.g. - 371 = 3^3 + 7^3 + 1^3. (3 digit number so power of 3).
Using this rationale, we find that
115132219018763992565095597973971522401 is the largest known armstrong number as per this and has 39 digits.
No wonder you forgot it!
Bonus reading - wikipedia
answered 2 days ago
RedBaronRedBaron
33626
New contributor
RedBaron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
answered 2 days ago
RedBaronRedBaron
33626
New contributor
RedBaron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
RedBaronRedBaron
33626
answered 2 days ago
RedBaronRedBaron
33626
33626
New contributor
RedBaron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
RedBaron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
RedBaron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
5
$begingroup$
Nice job! You got it! Side note: I actually used this (and all other password puzzles I post) as my password and I don't think I've forgotten one yet :) I like numbers!
$endgroup$
– Tyler22Alex
2 days ago
add a comment |
5
$begingroup$
Nice job! You got it! Side note: I actually used this (and all other password puzzles I post) as my password and I don't think I've forgotten one yet :) I like numbers!
$endgroup$
– Tyler22Alex
2 days ago
5
5
$begingroup$
Nice job! You got it! Side note: I actually used this (and all other password puzzles I post) as my password and I don't think I've forgotten one yet :) I like numbers!
$endgroup$
– Tyler22Alex
2 days ago
$begingroup$
Nice job! You got it! Side note: I actually used this (and all other password puzzles I post) as my password and I don't think I've forgotten one yet :) I like numbers!
$endgroup$
– Tyler22Alex
2 days ago
add a comment |
$begingroup$
I believe the answer is
1000
I think that we are supposed to take this literally as
"first of the word moon, which is M (1000 in Roman)"
answered Apr 2 at 19:49
AProughAPrough
5,9061245
$endgroup$
1
$begingroup$
A tag of the puzzle is knowledge. (ROT13) Oryvrir vg be abg, lbh pna gnxr bhg yrggref sebz "Ebzna Xabjyrqtr" arprffnel gb znxr gur jbeq "zbba". Creuncf lbh ner ba gur evtug genpx, ubjrire gur ahzore zvtug or n ovg fznyy nppbeqvat gb Uvag 1. $(+1)$ though!
$endgroup$
– user477343
Apr 3 at 0:36
1
$begingroup$
Very interesting answer, but not quite! Maybe take a look at the hints I added?
$endgroup$
– Tyler22Alex
Apr 3 at 0:44
$begingroup$
Yeah, answer was from before the hints. Nice puzzle though.
$endgroup$
– APrough
2 days ago
add a comment |
$begingroup$
I believe the answer is
1000
I think that we are supposed to take this literally as
"first of the word moon, which is M (1000 in Roman)"
answered Apr 2 at 19:49
AProughAPrough
5,9061245
$endgroup$
1
$begingroup$
A tag of the puzzle is knowledge. (ROT13) Oryvrir vg be abg, lbh pna gnxr bhg yrggref sebz "Ebzna Xabjyrqtr" arprffnel gb znxr gur jbeq "zbba". Creuncf lbh ner ba gur evtug genpx, ubjrire gur ahzore zvtug or n ovg fznyy nppbeqvat gb Uvag 1. $(+1)$ though!
$endgroup$
– user477343
Apr 3 at 0:36
1
$begingroup$
Very interesting answer, but not quite! Maybe take a look at the hints I added?
$endgroup$
– Tyler22Alex
Apr 3 at 0:44
$begingroup$
Yeah, answer was from before the hints. Nice puzzle though.
$endgroup$
– APrough
2 days ago
add a comment |
$begingroup$
I believe the answer is
1000
I think that we are supposed to take this literally as
"first of the word moon, which is M (1000 in Roman)"
answered Apr 2 at 19:49
AProughAPrough
5,9061245
$endgroup$
I believe the answer is
1000
I think that we are supposed to take this literally as
"first of the word moon, which is M (1000 in Roman)"
answered Apr 2 at 19:49
AProughAPrough
5,9061245
answered Apr 2 at 19:49
AProughAPrough
5,9061245
answered Apr 2 at 19:49
AProughAPrough
5,9061245
answered Apr 2 at 19:49
AProughAPrough
5,9061245
5,9061245
1
$begingroup$
A tag of the puzzle is knowledge. (ROT13) Oryvrir vg be abg, lbh pna gnxr bhg yrggref sebz "Ebzna Xabjyrqtr" arprffnel gb znxr gur jbeq "zbba". Creuncf lbh ner ba gur evtug genpx, ubjrire gur ahzore zvtug or n ovg fznyy nppbeqvat gb Uvag 1. $(+1)$ though!
$endgroup$
– user477343
Apr 3 at 0:36
1
$begingroup$
Very interesting answer, but not quite! Maybe take a look at the hints I added?
$endgroup$
– Tyler22Alex
Apr 3 at 0:44
$begingroup$
Yeah, answer was from before the hints. Nice puzzle though.
$endgroup$
– APrough
2 days ago
add a comment |
1
$begingroup$
A tag of the puzzle is knowledge. (ROT13) Oryvrir vg be abg, lbh pna gnxr bhg yrggref sebz "Ebzna Xabjyrqtr" arprffnel gb znxr gur jbeq "zbba". Creuncf lbh ner ba gur evtug genpx, ubjrire gur ahzore zvtug or n ovg fznyy nppbeqvat gb Uvag 1. $(+1)$ though!
$endgroup$
– user477343
Apr 3 at 0:36
1
$begingroup$
Very interesting answer, but not quite! Maybe take a look at the hints I added?
$endgroup$
– Tyler22Alex
Apr 3 at 0:44
$begingroup$
Yeah, answer was from before the hints. Nice puzzle though.
$endgroup$
– APrough
2 days ago
1
1
$begingroup$
A tag of the puzzle is knowledge. (ROT13) Oryvrir vg be abg, lbh pna gnxr bhg yrggref sebz "Ebzna Xabjyrqtr" arprffnel gb znxr gur jbeq "zbba". Creuncf lbh ner ba gur evtug genpx, ubjrire gur ahzore zvtug or n ovg fznyy nppbeqvat gb Uvag 1. $(+1)$ though!
$endgroup$
– user477343
Apr 3 at 0:36
$begingroup$
A tag of the puzzle is knowledge. (ROT13) Oryvrir vg be abg, lbh pna gnxr bhg yrggref sebz "Ebzna Xabjyrqtr" arprffnel gb znxr gur jbeq "zbba". Creuncf lbh ner ba gur evtug genpx, ubjrire gur ahzore zvtug or n ovg fznyy nppbeqvat gb Uvag 1. $(+1)$ though!
$endgroup$
– user477343
Apr 3 at 0:36
1
1
$begingroup$
Very interesting answer, but not quite! Maybe take a look at the hints I added?
$endgroup$
– Tyler22Alex
Apr 3 at 0:44
$begingroup$
Very interesting answer, but not quite! Maybe take a look at the hints I added?
$endgroup$
– Tyler22Alex
Apr 3 at 0:44
$begingroup$
Yeah, answer was from before the hints. Nice puzzle though.
$endgroup$
– APrough
2 days ago
$begingroup$
Yeah, answer was from before the hints. Nice puzzle though.
$endgroup$
– APrough
2 days ago
add a comment |
$begingroup$
Okay, so first of all, I checked the image and contrast to see if there was something hidden there. I couldn't find anything. Next, I followed the literal interpratation of the text, that is:
The first man to walk on the moon was Neil Armstrong. I tried looking for any items that he wore that would have a serial number or something like that. According to this site, https://www.fratellowatches.com/speedy-tuesday-speedmaster-professional-105-012-with-serial-number-2400xxxx/, the Speedwatches were the only ones used on the moon.
Neil Armstrong's serial number on his watch (again, according to the site) is 24002981, which I believe is your password.
answered Apr 2 at 17:25
Joe-You-KnowJoe-You-Know
6,59821073
$endgroup$
7
$begingroup$
As a number, you don't need people onboard to be first on the moon.
$endgroup$
– Arnaud Mortier
Apr 2 at 18:20
1
$begingroup$
On the right track with Armstrong, but looking in the wrong direction for the number. Maybe what you've got so far and my second hint will get you closer!
$endgroup$
– Tyler22Alex
2 days ago
add a comment |
$begingroup$
Okay, so first of all, I checked the image and contrast to see if there was something hidden there. I couldn't find anything. Next, I followed the literal interpratation of the text, that is:
The first man to walk on the moon was Neil Armstrong. I tried looking for any items that he wore that would have a serial number or something like that. According to this site, https://www.fratellowatches.com/speedy-tuesday-speedmaster-professional-105-012-with-serial-number-2400xxxx/, the Speedwatches were the only ones used on the moon.
Neil Armstrong's serial number on his watch (again, according to the site) is 24002981, which I believe is your password.
answered Apr 2 at 17:25
Joe-You-KnowJoe-You-Know
6,59821073
$endgroup$
7
$begingroup$
As a number, you don't need people onboard to be first on the moon.
$endgroup$
– Arnaud Mortier
Apr 2 at 18:20
1
$begingroup$
On the right track with Armstrong, but looking in the wrong direction for the number. Maybe what you've got so far and my second hint will get you closer!
$endgroup$
– Tyler22Alex
2 days ago
add a comment |
$begingroup$
Okay, so first of all, I checked the image and contrast to see if there was something hidden there. I couldn't find anything. Next, I followed the literal interpratation of the text, that is:
The first man to walk on the moon was Neil Armstrong. I tried looking for any items that he wore that would have a serial number or something like that. According to this site, https://www.fratellowatches.com/speedy-tuesday-speedmaster-professional-105-012-with-serial-number-2400xxxx/, the Speedwatches were the only ones used on the moon.
Neil Armstrong's serial number on his watch (again, according to the site) is 24002981, which I believe is your password.
answered Apr 2 at 17:25
Joe-You-KnowJoe-You-Know
6,59821073
$endgroup$
Okay, so first of all, I checked the image and contrast to see if there was something hidden there. I couldn't find anything. Next, I followed the literal interpratation of the text, that is:
The first man to walk on the moon was Neil Armstrong. I tried looking for any items that he wore that would have a serial number or something like that. According to this site, https://www.fratellowatches.com/speedy-tuesday-speedmaster-professional-105-012-with-serial-number-2400xxxx/, the Speedwatches were the only ones used on the moon.
Neil Armstrong's serial number on his watch (again, according to the site) is 24002981, which I believe is your password.
answered Apr 2 at 17:25
Joe-You-KnowJoe-You-Know
6,59821073
answered Apr 2 at 17:25
Joe-You-KnowJoe-You-Know
6,59821073
answered Apr 2 at 17:25
Joe-You-KnowJoe-You-Know
6,59821073
answered Apr 2 at 17:25
Joe-You-KnowJoe-You-Know
6,59821073
6,59821073
7
$begingroup$
As a number, you don't need people onboard to be first on the moon.
$endgroup$
– Arnaud Mortier
Apr 2 at 18:20
1
$begingroup$
On the right track with Armstrong, but looking in the wrong direction for the number. Maybe what you've got so far and my second hint will get you closer!
$endgroup$
– Tyler22Alex
2 days ago
add a comment |
7
$begingroup$
As a number, you don't need people onboard to be first on the moon.
$endgroup$
– Arnaud Mortier
Apr 2 at 18:20
1
$begingroup$
On the right track with Armstrong, but looking in the wrong direction for the number. Maybe what you've got so far and my second hint will get you closer!
$endgroup$
– Tyler22Alex
2 days ago
7
7
$begingroup$
As a number, you don't need people onboard to be first on the moon.
$endgroup$
– Arnaud Mortier
Apr 2 at 18:20
$begingroup$
As a number, you don't need people onboard to be first on the moon.
$endgroup$
– Arnaud Mortier
Apr 2 at 18:20
1
1
$begingroup$
On the right track with Armstrong, but looking in the wrong direction for the number. Maybe what you've got so far and my second hint will get you closer!
$endgroup$
– Tyler22Alex
2 days ago
$begingroup$
On the right track with Armstrong, but looking in the wrong direction for the number. Maybe what you've got so far and my second hint will get you closer!
$endgroup$
– Tyler22Alex
2 days ago
add a comment |
$begingroup$
Perhaps the password is
1959
Explanation:
Luna 2 was the first spacecraft to make contact with the moon.
According to Wikipedia, Luna 2 was engraved with "1959 январь" and "СОЮЗ СОВЕТСКИХ СОЦИАЛИСТИЧЕСКИХ РЕСПУБЛИК"1, making "1959" the first number on the moon.
Furthermore, phone passcodes are typically 4 digits long, so using a year as a passcode makes sense.
answered Apr 2 at 21:21
user1812user1812
912
$endgroup$
1
$begingroup$
Pursuant to Hint 1 however, this does not quite match, thus making it unlikely to be the answer; nevertheless, I like it! $(+1)$ :P
$endgroup$
– user477343
Apr 3 at 0:28
1
$begingroup$
I think this was posted before I added the first hint, so I apologize for that, but good thinking nonetheless!
$endgroup$
– Tyler22Alex
2 days ago
$begingroup$
@Tyler22Alex oh, it was? My bad. (Also, is it nevertheless or nonetheless...? Or are they interchangeable?)
$endgroup$
– user477343
2 days ago
add a comment |
$begingroup$
Perhaps the password is
1959
Explanation:
Luna 2 was the first spacecraft to make contact with the moon.
According to Wikipedia, Luna 2 was engraved with "1959 январь" and "СОЮЗ СОВЕТСКИХ СОЦИАЛИСТИЧЕСКИХ РЕСПУБЛИК"1, making "1959" the first number on the moon.
Furthermore, phone passcodes are typically 4 digits long, so using a year as a passcode makes sense.
answered Apr 2 at 21:21
user1812user1812
912
$endgroup$
1
$begingroup$
Pursuant to Hint 1 however, this does not quite match, thus making it unlikely to be the answer; nevertheless, I like it! $(+1)$ :P
$endgroup$
– user477343
Apr 3 at 0:28
1
$begingroup$
I think this was posted before I added the first hint, so I apologize for that, but good thinking nonetheless!
$endgroup$
– Tyler22Alex
2 days ago
$begingroup$
@Tyler22Alex oh, it was? My bad. (Also, is it nevertheless or nonetheless...? Or are they interchangeable?)
$endgroup$
– user477343
2 days ago
add a comment |
$begingroup$
Perhaps the password is
1959
Explanation:
Luna 2 was the first spacecraft to make contact with the moon.
According to Wikipedia, Luna 2 was engraved with "1959 январь" and "СОЮЗ СОВЕТСКИХ СОЦИАЛИСТИЧЕСКИХ РЕСПУБЛИК"1, making "1959" the first number on the moon.
Furthermore, phone passcodes are typically 4 digits long, so using a year as a passcode makes sense.
answered Apr 2 at 21:21
user1812user1812
912
$endgroup$
Perhaps the password is
1959
Explanation:
Luna 2 was the first spacecraft to make contact with the moon.
According to Wikipedia, Luna 2 was engraved with "1959 январь" and "СОЮЗ СОВЕТСКИХ СОЦИАЛИСТИЧЕСКИХ РЕСПУБЛИК"1, making "1959" the first number on the moon.
Furthermore, phone passcodes are typically 4 digits long, so using a year as a passcode makes sense.
answered Apr 2 at 21:21
user1812user1812
912
answered Apr 2 at 21:21
user1812user1812
912
answered Apr 2 at 21:21
user1812user1812
912
answered Apr 2 at 21:21
user1812user1812
912
912
1
$begingroup$
Pursuant to Hint 1 however, this does not quite match, thus making it unlikely to be the answer; nevertheless, I like it! $(+1)$ :P
$endgroup$
– user477343
Apr 3 at 0:28
1
$begingroup$
I think this was posted before I added the first hint, so I apologize for that, but good thinking nonetheless!
$endgroup$
– Tyler22Alex
2 days ago
$begingroup$
@Tyler22Alex oh, it was? My bad. (Also, is it nevertheless or nonetheless...? Or are they interchangeable?)
$endgroup$
– user477343
2 days ago
add a comment |
1
$begingroup$
Pursuant to Hint 1 however, this does not quite match, thus making it unlikely to be the answer; nevertheless, I like it! $(+1)$ :P
$endgroup$
– user477343
Apr 3 at 0:28
1
$begingroup$
I think this was posted before I added the first hint, so I apologize for that, but good thinking nonetheless!
$endgroup$
– Tyler22Alex
2 days ago
$begingroup$
@Tyler22Alex oh, it was? My bad. (Also, is it nevertheless or nonetheless...? Or are they interchangeable?)
$endgroup$
– user477343
2 days ago
1
1
$begingroup$
Pursuant to Hint 1 however, this does not quite match, thus making it unlikely to be the answer; nevertheless, I like it! $(+1)$ :P
$endgroup$
– user477343
Apr 3 at 0:28
$begingroup$
Pursuant to Hint 1 however, this does not quite match, thus making it unlikely to be the answer; nevertheless, I like it! $(+1)$ :P
$endgroup$
– user477343
Apr 3 at 0:28
1
1
$begingroup$
I think this was posted before I added the first hint, so I apologize for that, but good thinking nonetheless!
$endgroup$
– Tyler22Alex
2 days ago
$begingroup$
I think this was posted before I added the first hint, so I apologize for that, but good thinking nonetheless!
$endgroup$
– Tyler22Alex
2 days ago
$begingroup$
@Tyler22Alex oh, it was? My bad. (Also, is it nevertheless or nonetheless...? Or are they interchangeable?)
$endgroup$
– user477343
2 days ago
$begingroup$
@Tyler22Alex oh, it was? My bad. (Also, is it nevertheless or nonetheless...? Or are they interchangeable?)
$endgroup$
– user477343
2 days ago
add a comment |
$begingroup$
Perhaps the password is
11 or eleven or some variation of it
because
Apollo 11 was the first manned mission to the moon, and it had the “largest number” of anything that was first to the moon, greater that Luna 2, which was the first successful unmanned spacecraft to land.
answered Apr 2 at 19:07
PiIsNot3PiIsNot3
1,61222
$endgroup$
1
$begingroup$
I see the number in Apollo! $(+1)$
$endgroup$
– user477343
Apr 3 at 0:37
1
$begingroup$
Interesting idea! However I think you posted this before I added the first hint, so I apologize for that.
$endgroup$
– Tyler22Alex
2 days ago
add a comment |
$begingroup$
Perhaps the password is
11 or eleven or some variation of it
because
Apollo 11 was the first manned mission to the moon, and it had the “largest number” of anything that was first to the moon, greater that Luna 2, which was the first successful unmanned spacecraft to land.
answered Apr 2 at 19:07
PiIsNot3PiIsNot3
1,61222
$endgroup$
1
$begingroup$
I see the number in Apollo! $(+1)$
$endgroup$
– user477343
Apr 3 at 0:37
1
$begingroup$
Interesting idea! However I think you posted this before I added the first hint, so I apologize for that.
$endgroup$
– Tyler22Alex
2 days ago
add a comment |
$begingroup$
Perhaps the password is
11 or eleven or some variation of it
because
Apollo 11 was the first manned mission to the moon, and it had the “largest number” of anything that was first to the moon, greater that Luna 2, which was the first successful unmanned spacecraft to land.
answered Apr 2 at 19:07
PiIsNot3PiIsNot3
1,61222
$endgroup$
Perhaps the password is
11 or eleven or some variation of it
because
Apollo 11 was the first manned mission to the moon, and it had the “largest number” of anything that was first to the moon, greater that Luna 2, which was the first successful unmanned spacecraft to land.
answered Apr 2 at 19:07
PiIsNot3PiIsNot3
1,61222
answered Apr 2 at 19:07
PiIsNot3PiIsNot3
1,61222
answered Apr 2 at 19:07
PiIsNot3PiIsNot3
1,61222
answered Apr 2 at 19:07
PiIsNot3PiIsNot3
1,61222
1,61222
1
$begingroup$
I see the number in Apollo! $(+1)$
$endgroup$
– user477343
Apr 3 at 0:37
1
$begingroup$
Interesting idea! However I think you posted this before I added the first hint, so I apologize for that.
$endgroup$
– Tyler22Alex
2 days ago
add a comment |
1
$begingroup$
I see the number in Apollo! $(+1)$
$endgroup$
– user477343
Apr 3 at 0:37
1
$begingroup$
Interesting idea! However I think you posted this before I added the first hint, so I apologize for that.
$endgroup$
– Tyler22Alex
2 days ago
1
1
$begingroup$
I see the number in Apollo! $(+1)$
$endgroup$
– user477343
Apr 3 at 0:37
$begingroup$
I see the number in Apollo! $(+1)$
$endgroup$
– user477343
Apr 3 at 0:37
1
1
$begingroup$
Interesting idea! However I think you posted this before I added the first hint, so I apologize for that.
$endgroup$
– Tyler22Alex
2 days ago
$begingroup$
Interesting idea! However I think you posted this before I added the first hint, so I apologize for that.
$endgroup$
– Tyler22Alex
2 days ago
add a comment |
$begingroup$
There are a lot of ways to find numbers that are "first on the moon", so this question may need some more specificity to rule out answers; right now there doesn't seem to be any criteria beyond "the largest" number an answerer might justify somehow, which is probably too broad.
It's possible the password is
201901030226
We know that
OP uses passwords that are usually pretty long, as we learned in Unlock My Phone! January 2018.
We're looking for the largest "first on the moon" number.
China recently made the first landing ever on the far side of the moon - the robotic
Chang'e 4 mission touched down in the Von Kármán Crater at 10:26am Jan. 3, Beijing time.
This is clearly a "first" on the moon, both for China (first moon landing at all) and for humanity (first landing on the dark side of the moon).
If we take the date and time, in GMT, of this moon landing, we get 201901030226. Seconds were not given, so this is the most specific timedate stamp we can get.
Since this is the most recent moon landing, if we were to consider any other "first" - by nation, by mission type, whatever - and take their corresponding YYYYMMDDHMM number, this one would be the largest.
answered Apr 2 at 22:14
Rubio♦Rubio
30.5k567188
$endgroup$
1
$begingroup$
Hopefully my hint helped a bit, and I'll add more hints as time goes by if people still aren't getting it.
$endgroup$
– Tyler22Alex
2 days ago
add a comment |
$begingroup$
There are a lot of ways to find numbers that are "first on the moon", so this question may need some more specificity to rule out answers; right now there doesn't seem to be any criteria beyond "the largest" number an answerer might justify somehow, which is probably too broad.
It's possible the password is
201901030226
We know that
OP uses passwords that are usually pretty long, as we learned in Unlock My Phone! January 2018.
We're looking for the largest "first on the moon" number.
China recently made the first landing ever on the far side of the moon - the robotic
Chang'e 4 mission touched down in the Von Kármán Crater at 10:26am Jan. 3, Beijing time.
This is clearly a "first" on the moon, both for China (first moon landing at all) and for humanity (first landing on the dark side of the moon).
If we take the date and time, in GMT, of this moon landing, we get 201901030226. Seconds were not given, so this is the most specific timedate stamp we can get.
Since this is the most recent moon landing, if we were to consider any other "first" - by nation, by mission type, whatever - and take their corresponding YYYYMMDDHMM number, this one would be the largest.
answered Apr 2 at 22:14
Rubio♦Rubio
30.5k567188
$endgroup$
1
$begingroup$
Hopefully my hint helped a bit, and I'll add more hints as time goes by if people still aren't getting it.
$endgroup$
– Tyler22Alex
2 days ago
add a comment |
$begingroup$
There are a lot of ways to find numbers that are "first on the moon", so this question may need some more specificity to rule out answers; right now there doesn't seem to be any criteria beyond "the largest" number an answerer might justify somehow, which is probably too broad.
It's possible the password is
201901030226
We know that
OP uses passwords that are usually pretty long, as we learned in Unlock My Phone! January 2018.
We're looking for the largest "first on the moon" number.
China recently made the first landing ever on the far side of the moon - the robotic
Chang'e 4 mission touched down in the Von Kármán Crater at 10:26am Jan. 3, Beijing time.
This is clearly a "first" on the moon, both for China (first moon landing at all) and for humanity (first landing on the dark side of the moon).
If we take the date and time, in GMT, of this moon landing, we get 201901030226. Seconds were not given, so this is the most specific timedate stamp we can get.
Since this is the most recent moon landing, if we were to consider any other "first" - by nation, by mission type, whatever - and take their corresponding YYYYMMDDHMM number, this one would be the largest.
answered Apr 2 at 22:14
Rubio♦Rubio
30.5k567188
$endgroup$
There are a lot of ways to find numbers that are "first on the moon", so this question may need some more specificity to rule out answers; right now there doesn't seem to be any criteria beyond "the largest" number an answerer might justify somehow, which is probably too broad.
It's possible the password is
201901030226
We know that
OP uses passwords that are usually pretty long, as we learned in Unlock My Phone! January 2018.
We're looking for the largest "first on the moon" number.
China recently made the first landing ever on the far side of the moon - the robotic
Chang'e 4 mission touched down in the Von Kármán Crater at 10:26am Jan. 3, Beijing time.
This is clearly a "first" on the moon, both for China (first moon landing at all) and for humanity (first landing on the dark side of the moon).
If we take the date and time, in GMT, of this moon landing, we get 201901030226. Seconds were not given, so this is the most specific timedate stamp we can get.
Since this is the most recent moon landing, if we were to consider any other "first" - by nation, by mission type, whatever - and take their corresponding YYYYMMDDHMM number, this one would be the largest.
answered Apr 2 at 22:14
Rubio♦Rubio
30.5k567188
answered Apr 2 at 22:14
Rubio♦Rubio
30.5k567188
answered Apr 2 at 22:14
Rubio♦Rubio
30.5k567188
answered Apr 2 at 22:14
Rubio♦Rubio
30.5k567188
30.5k567188
1
$begingroup$
Hopefully my hint helped a bit, and I'll add more hints as time goes by if people still aren't getting it.
$endgroup$
– Tyler22Alex
2 days ago
add a comment |
1
$begingroup$
Hopefully my hint helped a bit, and I'll add more hints as time goes by if people still aren't getting it.
$endgroup$
– Tyler22Alex
2 days ago
1
1
$begingroup$
Hopefully my hint helped a bit, and I'll add more hints as time goes by if people still aren't getting it.
$endgroup$
– Tyler22Alex
2 days ago
$begingroup$
Hopefully my hint helped a bit, and I'll add more hints as time goes by if people still aren't getting it.
$endgroup$
– Tyler22Alex
2 days ago
add a comment |
$begingroup$
I think it's:
23.9 billion
Explanation:
According to Wikipedia:
In January 1969, NASA prepared an itemized estimate of the run-out cost of the Apollo program. The total came to $23.9 billion
answered Apr 2 at 21:05
tuvokkituvokki
1312
New contributor
tuvokki is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Good thinking but not quite! Maybe look at the hints I added. Good luck!
$endgroup$
– Tyler22Alex
2 days ago
add a comment |
$begingroup$
I think it's:
23.9 billion
Explanation:
According to Wikipedia:
In January 1969, NASA prepared an itemized estimate of the run-out cost of the Apollo program. The total came to $23.9 billion
answered Apr 2 at 21:05
tuvokkituvokki
1312
New contributor
tuvokki is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Good thinking but not quite! Maybe look at the hints I added. Good luck!
$endgroup$
– Tyler22Alex
2 days ago
add a comment |
$begingroup$
I think it's:
23.9 billion
Explanation:
According to Wikipedia:
In January 1969, NASA prepared an itemized estimate of the run-out cost of the Apollo program. The total came to $23.9 billion
answered Apr 2 at 21:05
tuvokkituvokki
1312
New contributor
tuvokki is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I think it's:
23.9 billion
Explanation:
According to Wikipedia:
In January 1969, NASA prepared an itemized estimate of the run-out cost of the Apollo program. The total came to $23.9 billion
answered Apr 2 at 21:05
tuvokkituvokki
1312
New contributor
tuvokki is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Apr 2 at 21:05
tuvokkituvokki
1312
New contributor
tuvokki is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Apr 2 at 21:05
tuvokkituvokki
1312
answered Apr 2 at 21:05
tuvokkituvokki
1312
1312
New contributor
tuvokki is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
tuvokki is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
tuvokki is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Good thinking but not quite! Maybe look at the hints I added. Good luck!
$endgroup$
– Tyler22Alex
2 days ago
add a comment |
1
$begingroup$
Good thinking but not quite! Maybe look at the hints I added. Good luck!
$endgroup$
– Tyler22Alex
2 days ago
1
1
$begingroup$
Good thinking but not quite! Maybe look at the hints I added. Good luck!
$endgroup$
– Tyler22Alex
2 days ago
$begingroup$
Good thinking but not quite! Maybe look at the hints I added. Good luck!
$endgroup$
– Tyler22Alex
2 days ago
add a comment |
$begingroup$
The answer is
9.5 Because it's the boot size Neil Armstrong had when first setting foot on the moon.
Thus it was "the largest number to be first on the moon"
answered 2 days ago
PeterPeter
2006
$endgroup$
add a comment |
$begingroup$
The answer is
9.5 Because it's the boot size Neil Armstrong had when first setting foot on the moon.
Thus it was "the largest number to be first on the moon"
answered 2 days ago
PeterPeter
2006
$endgroup$
add a comment |
$begingroup$
The answer is
9.5 Because it's the boot size Neil Armstrong had when first setting foot on the moon.
Thus it was "the largest number to be first on the moon"
answered 2 days ago
PeterPeter
2006
$endgroup$
The answer is
9.5 Because it's the boot size Neil Armstrong had when first setting foot on the moon.
Thus it was "the largest number to be first on the moon"
answered 2 days ago
PeterPeter
2006
answered 2 days ago
PeterPeter
2006
answered 2 days ago
PeterPeter
2006
answered 2 days ago
PeterPeter
2006
2006
add a comment |
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
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As a small recommendation, linked with Rubio's answer, if the January 2018 puzzle's hint also applies to this question, it would be better to also place it. It's better to get a repeated hint than a missing yet important hint
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– Belhenix
Apr 2 at 22:30
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@Belhenix I added that hint as well as one more. I'll keep this in mind for all future puzzles too, thanks!
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– Tyler22Alex
Apr 3 at 0:32