Which dates contain no repetitions of a digit?

Which dates contain no repetitions of a digit?

Question

This question concerns calendar dates of the form $d_1d_2/m_1m_2/y_1y_2y_3y_4$ in the order day/month/year.

The question specifically concerns those dates which contain no repetitions of a digit. For example, the date $23/05/1967$ is such a date but $07/12/1974$ is not such a date as both $1=m_1=y_1$ and $7=d_2=y_3$ are repeated digits.

We will use the Gregorian Calendar throughout (this is the calendar system that is standard throughout most of the world; see below.)

1. Show that there is no date with no repetition of digits in the years from $2000$ to $2099$.

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1. What was the last date before today, $03/11/2010$, with no repetition of digits? Explain your answer.

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1. When will the next such date be? Explain your answer.

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1. How many such dates were there in years from $1900$ to $1999$? Explain your answer.

[The Gregorian Calendar uses 12 months, which have, respectively, $31$, $28$ or $29$, $31$, $30, 31, 30, 31, 31, 30, 31, 30$ and $31$ days. The second month (February) has $28$ days in years that are not divisible by $4$, or that are divisible by $100$ but not $400$ (such as $1900$); it has $29$ days in the other years (leap years).]

Suggestion

This question concerns calendar dates of the form $d_1d_2/m_1m_2/y_1y_2y_3y_4$ in the order day/month/year.

The question specifically concerns those dates which contain no repetitions of a digit. For example, the date $23/05/1967$ is such date but $07/12/1974$ is not such a date as both $1=m_1=y_1$ and $7=d_2=y_3$ are repeated digits.

We will use the Gregorian Calendar throughout (this is the calendar system that is standard throughout most of the world; see below.)

Which digits are already in use?

Solution

This question concerns calendar dates of the form $d_1d_2/m_1m_2/y_1y_2y_3y_4$ in the order day/month/year.

The question specifically concerns those dates which contain no repetitions of a digit. For example, the date $23/05/1967$ is such date but $07/12/1974$ is not such a date as both $1=m_1=y_1$ and $7=d_2=y_3$ are repeated digits.

We will use the Gregorian Calendar throughout (this is the calendar system that is standard throughout most of the world; see below.)

Any date which has a year from $2000$ to $2099$ will have $y_1=2$ and $y_2=0$.

Therefore, $m_1$ cannot be $0$, so $m_1=1$.

Since there are only $12$ months, $m_2$ can be only $0$, $1$ or $2$, but any of these values will lead to repetitions.

Hence, there is no date with no repetition of digits in the years from $2000$ to $2099$.

We should aim for dates of the form $d_1d_2/m_1m_2/19y_3y_4$, since we already know that there is no date from $2000$ to $2099$ that will work.

Also, since we want the last day, we should find the biggest values possible, starting with the year. The greatest value for a year with no repeated digits is $1987$.

Now we aim to find the greatest possible value for the month. Since $10$,$11$ and $12$ will all lead to repeated digit $1$, this is $06$.

For the day, we cannot use $0$ and $1$ and so the day cannot be $30$ or $31$. So the biggest day with no repeated values is $25$.

Hence the last date before $03/11/2010$ with no repetition of digits is $25/06/1987$.

To find the next such date after $03/11/2010$ we aim to choose the smallest values for day, year and month, starting with the year.

Since every date from $2000$ to $2099$ repeats digits, the date we seek should be of the form $d_1d_2/m_1m_2/2y_2y_3y_4$, with $y_2>0$.

If $y_2=1$, then, in order to avoid repetitions, $m_1=0$ and $d_1=3$.

But then we will have no possible value for $d_2$, as we cannot use $0$ or $1$.

So $y_2$ cannot be $1$. Moreover, by the same argument, $y_3$ and $y_4$ should also be different from $1$.

The next possible value is $y_2=3$. Since we want the minimum value of $y_1y_2y_3y_4$, and we cannot use $1$ or $0$, $y_1y_2y_3y_4=2345$.

So the minimum value of the month that we can choose is $06$, and the minimum one for the day is $17$.

Therefore the next date with no repeated digits is $17/06/2345$.

Since we want years from $1900$ to $1999$, $y_1=1$ and $y_2=9$.

Also, $m_1=0$, as it cannot be $1$. Since we have used $1$ and $0$, and the only days starting with $3$ are $30$ and $31$, which cannot be used, $d_1=2$.

For the other positions, namely, $d_2$, $m_2$, $y_3$ and $y_4$, we can choose any of the $6$ digits that remain unused.

So we have $6$ ways of choosing $d_2$, $5$ ways of choosing $m_2$ (as it should be different from $d_2$), $4$ ways of choosing $y_3$ and $3$ ways of choosing $y_4$.

Hence, there are $6\times5\times4\times3=360$ such dates in the years from $1900$ to $1999$.