Divisible Dates
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The day of the month, and the month of the year, often simultaneously divide the year. Most recently it happened on January 3, 2019 because both 1 (January) and 3 divide 2019.
In our era, since 1/1/1, which years have had the most such dates?
arithmetic
asked Jan 25 at 14:56
Bernardo Recamán SantosBernardo Recamán Santos
2,6401346
$endgroup$
add a comment |
$begingroup$
The day of the month, and the month of the year, often simultaneously divide the year. Most recently it happened on January 3, 2019 because both 1 (January) and 3 divide 2019.
In our era, since 1/1/1, which years have had the most such dates?
arithmetic
asked Jan 25 at 14:56
Bernardo Recamán SantosBernardo Recamán Santos
2,6401346
$endgroup$
$begingroup$
For clarification: Is a divisible date one in which either the day or month divide the year, or one in with both do?
$endgroup$
– Van
Jan 25 at 15:11
$begingroup$
@Van Yes, both.
$endgroup$
– Bernardo Recamán Santos
Jan 25 at 15:16
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can we solve this with some code? :)
$endgroup$
– Flying_whale
Jan 25 at 15:17
$begingroup$
@Flying_whale: Go!
$endgroup$
– Bernardo Recamán Santos
Jan 25 at 15:18
add a comment |
$begingroup$
The day of the month, and the month of the year, often simultaneously divide the year. Most recently it happened on January 3, 2019 because both 1 (January) and 3 divide 2019.
In our era, since 1/1/1, which years have had the most such dates?
arithmetic
asked Jan 25 at 14:56
Bernardo Recamán SantosBernardo Recamán Santos
2,6401346
$endgroup$
The day of the month, and the month of the year, often simultaneously divide the year. Most recently it happened on January 3, 2019 because both 1 (January) and 3 divide 2019.
In our era, since 1/1/1, which years have had the most such dates?
arithmetic
arithmetic
asked Jan 25 at 14:56
Bernardo Recamán SantosBernardo Recamán Santos
2,6401346
asked Jan 25 at 14:56
Bernardo Recamán SantosBernardo Recamán Santos
2,6401346
edited Jan 26 at 14:14
Bernardo Recamán Santos
asked Jan 25 at 14:56
Bernardo Recamán SantosBernardo Recamán Santos
2,6401346
asked Jan 25 at 14:56
Bernardo Recamán SantosBernardo Recamán Santos
2,6401346
asked Jan 25 at 14:56
Bernardo Recamán SantosBernardo Recamán Santos
2,6401346
2,6401346
$begingroup$
For clarification: Is a divisible date one in which either the day or month divide the year, or one in with both do?
$endgroup$
– Van
Jan 25 at 15:11
$begingroup$
@Van Yes, both.
$endgroup$
– Bernardo Recamán Santos
Jan 25 at 15:16
$begingroup$
can we solve this with some code? :)
$endgroup$
– Flying_whale
Jan 25 at 15:17
$begingroup$
@Flying_whale: Go!
$endgroup$
– Bernardo Recamán Santos
Jan 25 at 15:18
add a comment |
$begingroup$
For clarification: Is a divisible date one in which either the day or month divide the year, or one in with both do?
$endgroup$
– Van
Jan 25 at 15:11
$begingroup$
@Van Yes, both.
$endgroup$
– Bernardo Recamán Santos
Jan 25 at 15:16
$begingroup$
can we solve this with some code? :)
$endgroup$
– Flying_whale
Jan 25 at 15:17
$begingroup$
@Flying_whale: Go!
$endgroup$
– Bernardo Recamán Santos
Jan 25 at 15:18
$begingroup$
For clarification: Is a divisible date one in which either the day or month divide the year, or one in with both do?
$endgroup$
– Van
Jan 25 at 15:11
$begingroup$
For clarification: Is a divisible date one in which either the day or month divide the year, or one in with both do?
$endgroup$
– Van
Jan 25 at 15:11
$begingroup$
@Van Yes, both.
$endgroup$
– Bernardo Recamán Santos
Jan 25 at 15:16
$begingroup$
@Van Yes, both.
$endgroup$
– Bernardo Recamán Santos
Jan 25 at 15:16
$begingroup$
can we solve this with some code? :)
$endgroup$
– Flying_whale
Jan 25 at 15:17
$begingroup$
can we solve this with some code? :)
$endgroup$
– Flying_whale
Jan 25 at 15:17
$begingroup$
@Flying_whale: Go!
$endgroup$
– Bernardo Recamán Santos
Jan 25 at 15:18
$begingroup$
@Flying_whale: Go!
$endgroup$
– Bernardo Recamán Santos
Jan 25 at 15:18
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
I'm going to go with
1680
factors:
1,2,3,4,5,6,7,8,10,12,14,15,16,20,21,24,28,30
So it hits on
179 days (10 months times 18 days, minus 30th of February)
Method:
Looked up
Highly composite numbers, picked the largest that's smaller than 2019.
answered Jan 25 at 15:39
BassBass
30.4k472186
$endgroup$
$begingroup$
Van got more days then this, so shouldn’t his be the accepted answer?
$endgroup$
– tyobrien
Jan 26 at 1:58
$begingroup$
@tyobrien If you look at the bottom right hand corner of the answer, you'll notice that this isn't the JonMark Perry answer Van managed to beat. :-)
$endgroup$
– Bass
Jan 26 at 2:25
add a comment |
$begingroup$
I found
with a bit of code i found 1680 with 179 dates you can run it here : https://rextester.com/YITA38880
answered Jan 25 at 15:33
Flying_whaleFlying_whale
1,909425
$endgroup$
add a comment |
$begingroup$
I believe I can beat JonMark Perry's answer:
1260 = 2 x2 x3 x3 x5 x7
For months, I can get 1, 2, 3, 4, 5, 6, 7, 9, 10, & 12 (10 - same).
But for days, I get:
1, 2, 3, 4, 5, 6, 7, 9, 10, 12 & 14, 15, 18, 20, 21, 28, 30 (17 - one more than Jon)
answered Jan 25 at 15:22
VanVan
59312
$endgroup$
add a comment |
$begingroup$
I'd be surprised if you can beat:
1080. It equals $2^3times3^3times5$, and so covers $1,2,3,4,5,6,8,9,10,12,15,18,20,24,27,30$ for days and $1,2,3,4,5,6,8,9,10,12$ for months.
answered Jan 25 at 15:10
JonMark PerryJonMark Perry
20.1k64098
$endgroup$
add a comment |
$begingroup$
This is a small enough set that a very simple program should be able to just brute-force it in a fraction of a second. So, here's a Python version. And, just to verify it, here's a wildly different one).1
If you run either one, the winner is:
1680, with 179 divisible days.
This makes perfect sense, for the reason Bass explains, but the brute-forcing shows how you could get the answer even without thinking very far.
That record obviously won't stand forever,2 but how soon will the record be broken?
Just change the outer loop to for year in itertools.count(1): to keep scanning until the end of time,3 and you'll see that:
2520 will have 208 divisible days.
1. Depending on the Python version, the datetime type may assume proleptic Georgian, so it'll incorrectly treat 400, 800, and 1200 as leap years. None of those three years is divisible by 29, so it won't make a difference. But better to actually get the rule right, as in the first version, than to get it wrong and then explain why it doesn't matter, right?
2. 2329089562800 is a leap year divisible by every number up to 31, so, assuming we use the Gregorian calendar that long, it will have 366 divisible dates.
3. Going all the way to 2329089562800 without optimizing the algorithm would probably take a while, but you can hit ^C after you get the first post-2019 record, or after you've seen enough that you're bored.
answered Jan 25 at 21:25
abarnertabarnert
1413
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I'm going to go with
1680
factors:
1,2,3,4,5,6,7,8,10,12,14,15,16,20,21,24,28,30
So it hits on
179 days (10 months times 18 days, minus 30th of February)
Method:
Looked up
Highly composite numbers, picked the largest that's smaller than 2019.
answered Jan 25 at 15:39
BassBass
30.4k472186
$endgroup$
$begingroup$
Van got more days then this, so shouldn’t his be the accepted answer?
$endgroup$
– tyobrien
Jan 26 at 1:58
$begingroup$
@tyobrien If you look at the bottom right hand corner of the answer, you'll notice that this isn't the JonMark Perry answer Van managed to beat. :-)
$endgroup$
– Bass
Jan 26 at 2:25
add a comment |
$begingroup$
I'm going to go with
1680
factors:
1,2,3,4,5,6,7,8,10,12,14,15,16,20,21,24,28,30
So it hits on
179 days (10 months times 18 days, minus 30th of February)
Method:
Looked up
Highly composite numbers, picked the largest that's smaller than 2019.
answered Jan 25 at 15:39
BassBass
30.4k472186
$endgroup$
$begingroup$
Van got more days then this, so shouldn’t his be the accepted answer?
$endgroup$
– tyobrien
Jan 26 at 1:58
$begingroup$
@tyobrien If you look at the bottom right hand corner of the answer, you'll notice that this isn't the JonMark Perry answer Van managed to beat. :-)
$endgroup$
– Bass
Jan 26 at 2:25
add a comment |
$begingroup$
I'm going to go with
1680
factors:
1,2,3,4,5,6,7,8,10,12,14,15,16,20,21,24,28,30
So it hits on
179 days (10 months times 18 days, minus 30th of February)
Method:
Looked up
Highly composite numbers, picked the largest that's smaller than 2019.
answered Jan 25 at 15:39
BassBass
30.4k472186
$endgroup$
I'm going to go with
1680
factors:
1,2,3,4,5,6,7,8,10,12,14,15,16,20,21,24,28,30
So it hits on
179 days (10 months times 18 days, minus 30th of February)
Method:
Looked up
Highly composite numbers, picked the largest that's smaller than 2019.
answered Jan 25 at 15:39
BassBass
30.4k472186
answered Jan 25 at 15:39
BassBass
30.4k472186
answered Jan 25 at 15:39
BassBass
30.4k472186
answered Jan 25 at 15:39
BassBass
30.4k472186
30.4k472186
$begingroup$
Van got more days then this, so shouldn’t his be the accepted answer?
$endgroup$
– tyobrien
Jan 26 at 1:58
$begingroup$
@tyobrien If you look at the bottom right hand corner of the answer, you'll notice that this isn't the JonMark Perry answer Van managed to beat. :-)
$endgroup$
– Bass
Jan 26 at 2:25
add a comment |
$begingroup$
Van got more days then this, so shouldn’t his be the accepted answer?
$endgroup$
– tyobrien
Jan 26 at 1:58
$begingroup$
@tyobrien If you look at the bottom right hand corner of the answer, you'll notice that this isn't the JonMark Perry answer Van managed to beat. :-)
$endgroup$
– Bass
Jan 26 at 2:25
$begingroup$
Van got more days then this, so shouldn’t his be the accepted answer?
$endgroup$
– tyobrien
Jan 26 at 1:58
$begingroup$
Van got more days then this, so shouldn’t his be the accepted answer?
$endgroup$
– tyobrien
Jan 26 at 1:58
$begingroup$
@tyobrien If you look at the bottom right hand corner of the answer, you'll notice that this isn't the JonMark Perry answer Van managed to beat. :-)
$endgroup$
– Bass
Jan 26 at 2:25
$begingroup$
@tyobrien If you look at the bottom right hand corner of the answer, you'll notice that this isn't the JonMark Perry answer Van managed to beat. :-)
$endgroup$
– Bass
Jan 26 at 2:25
add a comment |
$begingroup$
I found
with a bit of code i found 1680 with 179 dates you can run it here : https://rextester.com/YITA38880
answered Jan 25 at 15:33
Flying_whaleFlying_whale
1,909425
$endgroup$
add a comment |
$begingroup$
I found
with a bit of code i found 1680 with 179 dates you can run it here : https://rextester.com/YITA38880
answered Jan 25 at 15:33
Flying_whaleFlying_whale
1,909425
$endgroup$
add a comment |
$begingroup$
I found
with a bit of code i found 1680 with 179 dates you can run it here : https://rextester.com/YITA38880
answered Jan 25 at 15:33
Flying_whaleFlying_whale
1,909425
$endgroup$
I found
with a bit of code i found 1680 with 179 dates you can run it here : https://rextester.com/YITA38880
answered Jan 25 at 15:33
Flying_whaleFlying_whale
1,909425
edited Jan 25 at 19:26
answered Jan 25 at 15:33
Flying_whaleFlying_whale
1,909425
answered Jan 25 at 15:33
Flying_whaleFlying_whale
1,909425
answered Jan 25 at 15:33
Flying_whaleFlying_whale
1,909425
1,909425
add a comment |
add a comment |
$begingroup$
I believe I can beat JonMark Perry's answer:
1260 = 2 x2 x3 x3 x5 x7
For months, I can get 1, 2, 3, 4, 5, 6, 7, 9, 10, & 12 (10 - same).
But for days, I get:
1, 2, 3, 4, 5, 6, 7, 9, 10, 12 & 14, 15, 18, 20, 21, 28, 30 (17 - one more than Jon)
answered Jan 25 at 15:22
VanVan
59312
$endgroup$
add a comment |
$begingroup$
I believe I can beat JonMark Perry's answer:
1260 = 2 x2 x3 x3 x5 x7
For months, I can get 1, 2, 3, 4, 5, 6, 7, 9, 10, & 12 (10 - same).
But for days, I get:
1, 2, 3, 4, 5, 6, 7, 9, 10, 12 & 14, 15, 18, 20, 21, 28, 30 (17 - one more than Jon)
answered Jan 25 at 15:22
VanVan
59312
$endgroup$
add a comment |
$begingroup$
I believe I can beat JonMark Perry's answer:
1260 = 2 x2 x3 x3 x5 x7
For months, I can get 1, 2, 3, 4, 5, 6, 7, 9, 10, & 12 (10 - same).
But for days, I get:
1, 2, 3, 4, 5, 6, 7, 9, 10, 12 & 14, 15, 18, 20, 21, 28, 30 (17 - one more than Jon)
answered Jan 25 at 15:22
VanVan
59312
$endgroup$
I believe I can beat JonMark Perry's answer:
1260 = 2 x2 x3 x3 x5 x7
For months, I can get 1, 2, 3, 4, 5, 6, 7, 9, 10, & 12 (10 - same).
But for days, I get:
1, 2, 3, 4, 5, 6, 7, 9, 10, 12 & 14, 15, 18, 20, 21, 28, 30 (17 - one more than Jon)
answered Jan 25 at 15:22
VanVan
59312
answered Jan 25 at 15:22
VanVan
59312
answered Jan 25 at 15:22
VanVan
59312
answered Jan 25 at 15:22
VanVan
59312
59312
add a comment |
add a comment |
$begingroup$
I'd be surprised if you can beat:
1080. It equals $2^3times3^3times5$, and so covers $1,2,3,4,5,6,8,9,10,12,15,18,20,24,27,30$ for days and $1,2,3,4,5,6,8,9,10,12$ for months.
answered Jan 25 at 15:10
JonMark PerryJonMark Perry
20.1k64098
$endgroup$
add a comment |
$begingroup$
I'd be surprised if you can beat:
1080. It equals $2^3times3^3times5$, and so covers $1,2,3,4,5,6,8,9,10,12,15,18,20,24,27,30$ for days and $1,2,3,4,5,6,8,9,10,12$ for months.
answered Jan 25 at 15:10
JonMark PerryJonMark Perry
20.1k64098
$endgroup$
add a comment |
$begingroup$
I'd be surprised if you can beat:
1080. It equals $2^3times3^3times5$, and so covers $1,2,3,4,5,6,8,9,10,12,15,18,20,24,27,30$ for days and $1,2,3,4,5,6,8,9,10,12$ for months.
answered Jan 25 at 15:10
JonMark PerryJonMark Perry
20.1k64098
$endgroup$
I'd be surprised if you can beat:
1080. It equals $2^3times3^3times5$, and so covers $1,2,3,4,5,6,8,9,10,12,15,18,20,24,27,30$ for days and $1,2,3,4,5,6,8,9,10,12$ for months.
answered Jan 25 at 15:10
JonMark PerryJonMark Perry
20.1k64098
answered Jan 25 at 15:10
JonMark PerryJonMark Perry
20.1k64098
answered Jan 25 at 15:10
JonMark PerryJonMark Perry
20.1k64098
answered Jan 25 at 15:10
JonMark PerryJonMark Perry
20.1k64098
20.1k64098
add a comment |
add a comment |
$begingroup$
This is a small enough set that a very simple program should be able to just brute-force it in a fraction of a second. So, here's a Python version. And, just to verify it, here's a wildly different one).1
If you run either one, the winner is:
1680, with 179 divisible days.
This makes perfect sense, for the reason Bass explains, but the brute-forcing shows how you could get the answer even without thinking very far.
That record obviously won't stand forever,2 but how soon will the record be broken?
Just change the outer loop to for year in itertools.count(1): to keep scanning until the end of time,3 and you'll see that:
2520 will have 208 divisible days.
1. Depending on the Python version, the datetime type may assume proleptic Georgian, so it'll incorrectly treat 400, 800, and 1200 as leap years. None of those three years is divisible by 29, so it won't make a difference. But better to actually get the rule right, as in the first version, than to get it wrong and then explain why it doesn't matter, right?
2. 2329089562800 is a leap year divisible by every number up to 31, so, assuming we use the Gregorian calendar that long, it will have 366 divisible dates.
3. Going all the way to 2329089562800 without optimizing the algorithm would probably take a while, but you can hit ^C after you get the first post-2019 record, or after you've seen enough that you're bored.
answered Jan 25 at 21:25
abarnertabarnert
1413
$endgroup$
add a comment |
$begingroup$
This is a small enough set that a very simple program should be able to just brute-force it in a fraction of a second. So, here's a Python version. And, just to verify it, here's a wildly different one).1
If you run either one, the winner is:
1680, with 179 divisible days.
This makes perfect sense, for the reason Bass explains, but the brute-forcing shows how you could get the answer even without thinking very far.
That record obviously won't stand forever,2 but how soon will the record be broken?
Just change the outer loop to for year in itertools.count(1): to keep scanning until the end of time,3 and you'll see that:
2520 will have 208 divisible days.
1. Depending on the Python version, the datetime type may assume proleptic Georgian, so it'll incorrectly treat 400, 800, and 1200 as leap years. None of those three years is divisible by 29, so it won't make a difference. But better to actually get the rule right, as in the first version, than to get it wrong and then explain why it doesn't matter, right?
2. 2329089562800 is a leap year divisible by every number up to 31, so, assuming we use the Gregorian calendar that long, it will have 366 divisible dates.
3. Going all the way to 2329089562800 without optimizing the algorithm would probably take a while, but you can hit ^C after you get the first post-2019 record, or after you've seen enough that you're bored.
answered Jan 25 at 21:25
abarnertabarnert
1413
$endgroup$
add a comment |
$begingroup$
This is a small enough set that a very simple program should be able to just brute-force it in a fraction of a second. So, here's a Python version. And, just to verify it, here's a wildly different one).1
If you run either one, the winner is:
1680, with 179 divisible days.
This makes perfect sense, for the reason Bass explains, but the brute-forcing shows how you could get the answer even without thinking very far.
That record obviously won't stand forever,2 but how soon will the record be broken?
Just change the outer loop to for year in itertools.count(1): to keep scanning until the end of time,3 and you'll see that:
2520 will have 208 divisible days.
1. Depending on the Python version, the datetime type may assume proleptic Georgian, so it'll incorrectly treat 400, 800, and 1200 as leap years. None of those three years is divisible by 29, so it won't make a difference. But better to actually get the rule right, as in the first version, than to get it wrong and then explain why it doesn't matter, right?
2. 2329089562800 is a leap year divisible by every number up to 31, so, assuming we use the Gregorian calendar that long, it will have 366 divisible dates.
3. Going all the way to 2329089562800 without optimizing the algorithm would probably take a while, but you can hit ^C after you get the first post-2019 record, or after you've seen enough that you're bored.
answered Jan 25 at 21:25
abarnertabarnert
1413
$endgroup$
This is a small enough set that a very simple program should be able to just brute-force it in a fraction of a second. So, here's a Python version. And, just to verify it, here's a wildly different one).1
If you run either one, the winner is:
1680, with 179 divisible days.
This makes perfect sense, for the reason Bass explains, but the brute-forcing shows how you could get the answer even without thinking very far.
That record obviously won't stand forever,2 but how soon will the record be broken?
Just change the outer loop to for year in itertools.count(1): to keep scanning until the end of time,3 and you'll see that:
2520 will have 208 divisible days.
1. Depending on the Python version, the datetime type may assume proleptic Georgian, so it'll incorrectly treat 400, 800, and 1200 as leap years. None of those three years is divisible by 29, so it won't make a difference. But better to actually get the rule right, as in the first version, than to get it wrong and then explain why it doesn't matter, right?
2. 2329089562800 is a leap year divisible by every number up to 31, so, assuming we use the Gregorian calendar that long, it will have 366 divisible dates.
3. Going all the way to 2329089562800 without optimizing the algorithm would probably take a while, but you can hit ^C after you get the first post-2019 record, or after you've seen enough that you're bored.
answered Jan 25 at 21:25
abarnertabarnert
1413
answered Jan 25 at 21:25
abarnertabarnert
1413
answered Jan 25 at 21:25
abarnertabarnert
1413
answered Jan 25 at 21:25
abarnertabarnert
1413
1413
add a comment |
add a comment |
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For clarification: Is a divisible date one in which either the day or month divide the year, or one in with both do?
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– Van
Jan 25 at 15:11
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@Van Yes, both.
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– Bernardo Recamán Santos
Jan 25 at 15:16
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can we solve this with some code? :)
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– Flying_whale
Jan 25 at 15:17
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@Flying_whale: Go!
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– Bernardo Recamán Santos
Jan 25 at 15:18