# Dates with 8 consecutive digits

In many places, dates are written as DD/MM/YYYY. For example, the 25th of April 1736 is written as 25/04/1736. Dates such as this one that use 8 consecutive digits (not necessarily in order) will be called illions.

1. What is the first illion after 2015?
2. Why must there be a 0 in every illion in the years 2000 to 2999?
3. Why must every illion in the years 2000 to 2999 have 0 as the first digit of the month?
4. How many illions are in the years 2000 to 2999?

(You cannot use the same digit twice in an illion)

I have tried to list all of the possibilities however, I have found this highly time consuming. I was wondering what method I should use to solve these questions without listing all the possibilities.

Thank you :)

• Welcome to Math.SE. It is customary to show what you have tried yourself to answer your question, so that other users can help you in a better way. Could you write about your efforts? – Hrodelbert Apr 14 '15 at 11:06
• Wow! This is a very interesting question. Anybody got any ideas? – anonymous Apr 14 '15 at 11:21
• Is it a requirement that you don't repeat digits? Could I express 12 November 2111 as 12/11/2111 which does not have a zero and is in the years 2000 to 2999? – Warren Hill Apr 14 '15 at 12:11
• You can't repeat digits because it says 8 consecutive digits. – Cool Guy Apr 14 '15 at 12:49
• @String 's answer is correct. I've tried implementing it programmatically (using brute force) and got the same answer. Here is the working fiddle if anyone wants to look at it. – Tachyon Apr 14 '15 at 18:04

(1) As argued in (2) below, the month contains a $0$ or is $12$, hence either the year is $>2999$ or does not contain a $0$. This makes $2134$ the earliest possible year. However, the leading digit of the day is $0$, $1$, $2$, or $3$, so that we must up the year to $2145$ at least, which allows day $30$ but then conflicts with the $0$ in the month. The next option is that the year does not use $1$ (nor $0$), which happens for the first time in $2345$. We can find an illion in that year: We know we use $0$ in the month (preferably as leading digit). Then the leading digit of the day must be $1$. Now the lowest valid month is $06$ and we can take $17/06/2345$

(2) If the two-digit month does not contain a zero and has two distinct digits, it must be December. But then the use of $2$ conflicts with the leading $2$ of the year. Actually, we see that the same argument applies to the year range $1000$ to $2999$.

As in (2), we know that the month cannot be $12$ or $11$. In cannot be $10$ either, because that makes $0,1,2$ used up by month and year, hence the day must have leading digit $3$, but neither $30$ nor $31$ are allowed

For (4), you may want to first enumerate valid day/month combinations (without $2$!) and look for years that fit the gaps.

• I am finding your explanation hard to understand and follow. – Cool Guy Apr 15 '15 at 12:27

1. Why must there be a 0 in every illion in the years 2000 to 2999?

Suppose that there exists an illion which does not have $0$. Let $AB/CD/2EFG$. Since $C$ has to be either $0$ or $1$, one has $C=1$. So, the $8$ consecutive numbers have to be $12345678$ (not $23456789$). But this is a contradiction because no number in $345678$ can fit for $D$.

1. Why must every illion in the years 2000 to 2999 have 0 as the first digit of the month?

Suppose that there exists an illion whose first digit of the month is $1$. This is because the first digit of the month has to be either $0$ or $1$. Let $AB/1C/2DEF$. One has $C=0$ because $C$ has to be either $0,1,2$ (Note that the first digit of the month is $1$ and that $1,2$ are already used). Also, one has $A=3$ because $A$ has to be either $0,1,2,3$ but $0,1,2$ are already used. However, this is a contradiction because no number in $34567$ can fit for $B$.

• Hmm. I still don't find this very satisfying for some reason. – anonymous Apr 14 '15 at 11:44
• @anonymous: I added some explanations. better? – mathlove Apr 14 '15 at 11:50
• Yes, that is much better. – anonymous Apr 14 '15 at 12:01

Assuming that we are done with $(1)-(3)$, number $(4)$ can be handled as follows:

We have $DD/0M/2YYY$.

For $DD=31$ we have $M\in\{5,7\}$ and the remaining digits of the year can be shuffled. This makes $2\cdot 3!=12$ dates.

For leading $D=1$, $M$ can be chosen freely. Thus the remaining five digits can be shuffled making $5!=120$ more dates.

Thus we have a total of $132$ possible dates, illions that is :o)

• @RossMillikan: You are right! Good point. – String Apr 14 '15 at 15:26
• @RossMillikan: I hope that I am done making stupid mistakes for now. Better? – String Apr 14 '15 at 15:27

for (d) there are 5 numbers to play around with in the 2nd day spot the 2nd month spot and the 2nd 3rd and 4rth year spots there are 5 different days and 4 different months and 6 different years per day. this means that there is 1560 illions between 2000 and 2999 which doesn't really make sense but any way that's the answer I got

• There's no such thing as "this result doesn't make sense, but it's the one I got". Right answers, if explained correctly, make sense. Wrong answers don't. Right answers are helpful, wrong ones aren't. – user228113 May 26 '15 at 8:20

0 to 7 are the lowest consecutive numbers you can use. Months only consist of 0,1,2. Use the 2,3,4,5 for the year. That is the next closest. Now your trying to find the shortest month so the 0 is used. Now you cant use the 1 here otherwise you wont be able to make a date up for the day. Eg 67/01/2345 is impossible. Therefore you use the lowest number as June is before July and the 7 goes to the day. 17/06/2345

• "Months only consist of 0,1,2" is false. "Use the 2,3,4,5 for the year. That is the next closest" needs to be justified. – epimorphic Jun 5 '15 at 23:36