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Next date with distinct digits

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5

$begingroup$

The date today is 7th June 2019, or 07/06/2019 (using the English DD/MM/YYYY ordering).

When is the next date that when written in this way has all eight digits different?

logical-deduction

share|improve this question

edited 7 hours ago

rnaylor

asked 9 hours ago

rnaylorrnaylor

1,2461921

$endgroup$

add a comment | 

5

$begingroup$

The date today is 7th June 2019, or 07/06/2019 (using the English DD/MM/YYYY ordering).

When is the next date that when written in this way has all eight digits different?

logical-deduction

share|improve this question

edited 7 hours ago

rnaylor

asked 9 hours ago

rnaylorrnaylor

1,2461921

$endgroup$

add a comment | 

$begingroup$

The date today is 7th June 2019, or 07/06/2019 (using the English DD/MM/YYYY ordering).

When is the next date that when written in this way has all eight digits different?

logical-deduction

share|improve this question

edited 7 hours ago

rnaylor

asked 9 hours ago

rnaylorrnaylor

1,2461921

$endgroup$

share|improve this question

edited 7 hours ago

rnaylor

asked 9 hours ago

rnaylorrnaylor

1,2461921

share|improve this question

edited 7 hours ago

rnaylor

asked 9 hours ago

rnaylorrnaylor

1,2461921

asked 9 hours ago

rnaylorrnaylor

1,2461921

asked 9 hours ago

rnaylorrnaylor

1,2461921

1 Answer
1

active

oldest

votes

8

$begingroup$

Rather remarkably, I wrote down this exact puzzle in my notebook a couple of years ago to which I think the answer is

17/06/2345 in DD/MM/YYYY format.

Reasoning

Notice that the first M will either be $0$ or $1$.
If it is $0$ then the first D will either be $1$ or $2$ or DD will be $31$.
If it is $1$ then either the second M will be $0$ or the second M will be $2$ and the day will contain a $0$.

Overall, this means that $0$ and either $1$ or $2$ must be used in the DD/MM part. If we don't want to skip to the next millenium, we need the $2$ for the beginning of the year.
Hence the DD/MM part requires both $0$ and $1$.
After that, we focus on the nearest year possible which comes from assigning the digits $3,4,5$ in order to century, decade and digit of the year.
It makes more sense to assign the $0$ to the month instead of $1$ but we cannot assign both since we cannot have a day without any of $0$, $1$ or $2$. Hence, we assign $6$ to the month and then $7$ to create the day.

share|improve this answer

edited 8 hours ago

answered 9 hours ago

hexominohexomino

52.4k4155247

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  I can confirm that this is also the result of writing a dumb Python program to try future dates until it finds one satisfying the given condition.
  $endgroup$
  – Gareth McCaughan♦
  7 hours ago
  

  
  
  
  

add a comment | 

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8

$begingroup$

Rather remarkably, I wrote down this exact puzzle in my notebook a couple of years ago to which I think the answer is

17/06/2345 in DD/MM/YYYY format.

Reasoning

Notice that the first M will either be $0$ or $1$.
If it is $0$ then the first D will either be $1$ or $2$ or DD will be $31$.
If it is $1$ then either the second M will be $0$ or the second M will be $2$ and the day will contain a $0$.

Overall, this means that $0$ and either $1$ or $2$ must be used in the DD/MM part. If we don't want to skip to the next millenium, we need the $2$ for the beginning of the year.
Hence the DD/MM part requires both $0$ and $1$.
After that, we focus on the nearest year possible which comes from assigning the digits $3,4,5$ in order to century, decade and digit of the year.
It makes more sense to assign the $0$ to the month instead of $1$ but we cannot assign both since we cannot have a day without any of $0$, $1$ or $2$. Hence, we assign $6$ to the month and then $7$ to create the day.

share|improve this answer

edited 8 hours ago

answered 9 hours ago

hexominohexomino

52.4k4155247

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  I can confirm that this is also the result of writing a dumb Python program to try future dates until it finds one satisfying the given condition.
  $endgroup$
  – Gareth McCaughan♦
  7 hours ago
  

  
  
  
  

add a comment | 

8

$begingroup$

Rather remarkably, I wrote down this exact puzzle in my notebook a couple of years ago to which I think the answer is

17/06/2345 in DD/MM/YYYY format.

Reasoning

Notice that the first M will either be $0$ or $1$.
If it is $0$ then the first D will either be $1$ or $2$ or DD will be $31$.
If it is $1$ then either the second M will be $0$ or the second M will be $2$ and the day will contain a $0$.

Overall, this means that $0$ and either $1$ or $2$ must be used in the DD/MM part. If we don't want to skip to the next millenium, we need the $2$ for the beginning of the year.
Hence the DD/MM part requires both $0$ and $1$.
After that, we focus on the nearest year possible which comes from assigning the digits $3,4,5$ in order to century, decade and digit of the year.
It makes more sense to assign the $0$ to the month instead of $1$ but we cannot assign both since we cannot have a day without any of $0$, $1$ or $2$. Hence, we assign $6$ to the month and then $7$ to create the day.

share|improve this answer

edited 8 hours ago

answered 9 hours ago

hexominohexomino

52.4k4155247

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  I can confirm that this is also the result of writing a dumb Python program to try future dates until it finds one satisfying the given condition.
  $endgroup$
  – Gareth McCaughan♦
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Rather remarkably, I wrote down this exact puzzle in my notebook a couple of years ago to which I think the answer is

17/06/2345 in DD/MM/YYYY format.

Reasoning

Notice that the first M will either be $0$ or $1$.
If it is $0$ then the first D will either be $1$ or $2$ or DD will be $31$.
If it is $1$ then either the second M will be $0$ or the second M will be $2$ and the day will contain a $0$.

Overall, this means that $0$ and either $1$ or $2$ must be used in the DD/MM part. If we don't want to skip to the next millenium, we need the $2$ for the beginning of the year.
Hence the DD/MM part requires both $0$ and $1$.
After that, we focus on the nearest year possible which comes from assigning the digits $3,4,5$ in order to century, decade and digit of the year.
It makes more sense to assign the $0$ to the month instead of $1$ but we cannot assign both since we cannot have a day without any of $0$, $1$ or $2$. Hence, we assign $6$ to the month and then $7$ to create the day.

share|improve this answer

edited 8 hours ago

answered 9 hours ago

hexominohexomino

52.4k4155247

$endgroup$

share|improve this answer

edited 8 hours ago

answered 9 hours ago

hexominohexomino

52.4k4155247

answered 9 hours ago

hexominohexomino

52.4k4155247

answered 9 hours ago

hexominohexomino

52.4k4155247