Smallest future date which involves no repetition of a digit in the format DD/MM/YYYY

My approach:

  • MM must contain at least one of $0,1$
  • DD must contain at least one of $0,1,2$
  • Hence YY can contain only one of $0,1,2$

Therefore:

  • Start with the smallest future year which contains only one of $[0,1,2]:2345$
  • Continue with the smallest month which contains only one of $[0,1]:06$
  • Finish with the smallest day which contains only one of $[1]:17$

17/06/2345 is the soonest such date.

Why?

  • xx/xx/20xx does not work. The month then cannot be 01 through 10 (the zero would be repeated). The month can never be 11. The month cannot be 12, either, because of the repetition of 2.

  • xx/xx/21xx does not work. The day then cannot start with 1 or 2. If it started with 0, then the month would have to begin with 1, but then 1 would be repeated. Therefore, the day would have to be either 30 or 31. 31 does not work (1 is repeated). If you put 30 as the day, then the month can start with neither 0 nor 1. This exhausts all the possibilities.

  • xx/xx/22xx obviously does not work.

  • Let us look at xx/xx/23xx. You may want the year to be 230x, but then the month must be 12, and 2 would be repeated. If the year is to be 231x, then the month has to be 04 through 09. But then the day could not start with 0, 1, 2, or 3. Therefore, 231x is out. 232x and 233x are obviously out. How about 234x? You may be tempted to try 2340, but then the month must be 12, which it cannot be. If the year is 2341, then the month has to be 05 through 09, but then the day could not start with 0, 1, 2, or 3. 2342 through 2344 obviously do not work. Hence, the soonest year can only be as far as 2345. I leave it to you to figure out the month and the day. [Hint: Why does 01 not work as the month?]


The first digit of the month must be $0$ or $1$. If it is $1$, the second digit must be $2$ or $0$; being $2$ would lead us to the fourth millennium, and being $0$ is useless because we could choose $01$ for the month instead.

So the first digit of the month must be $0$. The first digit of the year should be $2$, and the first digit of the day must be $1$ or $3$. But if it is $3$ the second digit of the day must be $1$ and year would be $\ge2400$. So, let's pick $1$ for the first digit of the day. We have this so far:

$$1D/0M/23YY$$

Can you finish?

Tags:

Recreational Mathematics

Elementary Number Theory

Calendar Computations

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