What is the smallest future date which involves no repetition of a digit in the format DD/MM/YYYY for the year? What is your approach?
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4Are single digit days/months allowed? Or is, say, July recorded as $07$ rather than just $7$? If so, the next such date is the day after tomorrow, July $8$th $8/7/2016$ – Rob Jul 06 '16 at 17:39
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5I might tell you on June 17th, 2345. :-) – quid Jul 06 '16 at 17:40
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1“[Elementary number theory]” my foot! More like [recreational mathematics]. – PJTraill Jul 06 '16 at 21:27
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4@PJTraill We do have the [calendar computations] tag... – David R. Jul 06 '16 at 21:52
4 Answers
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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 $[1]:17$
barak manos
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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?]
triple_sec
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101 does not work as the month. The day must not then start with 0 or 1, and must not start with 2 or 3, because of the year. – EKons Jul 07 '16 at 07:17
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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?
ajotatxe
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1Thanks. October can be excluded because it can be substituted by January with no more changes.. – ajotatxe Jul 06 '16 at 17:51
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If we require the date and month to use both digits without an initial zero, then we need to wait until October 27, 3456 (27/10/3456 Europe, 10/27/3456 USA). The constraints are: we must use 0, 1, 2 in the month and date (the month has two of those digits and the date can't have both digits greater than 2, then we use lowest available digit for the millenium, then the century, etc.
Oscar Lanzi
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