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### Closest Palindromic Dates

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Date: 02/07/2000 at 23:44:24
From: John Hessam
Subject: Palindromic dates of the 1900s (date.month.year)

Dear Dr. Math,

Problem: Using the abbreviation date.month.year (the last two digits
of the year), what are the 2 palindromic dates (of any number of
digits) closest together in the 1900s?

I have figured out that 1.1.11 and 11.1.11 are relatively close (10
days apart.) However, I have also found that 1.11.11 and 11.11.11 are
also ten days apart. I am wondering if there are any other two
palindromic dates in the 1900s closer together than ten days?

I'm extremely curious to find an approach other than "trial and error"
to solving this problem. Thanks.
```

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Date: 02/08/2000 at 17:20:10
From: Doctor TWE
Subject: Re: Palindromic dates of the 1900s (date.month.year)

Hi John - thanks for writing to Dr. Math.

That's a good question. I found a pair of palindromic dates that were
3 days apart. Here is an outline of my thinking:

In order to be less than 10 days apart, the two dates would either
have to fall in the same year, or to fall in December of one year and
January of the next. I'll consider the latter first; it's easier.

If the earlier date falls in December of some year, it must be of one
of the forms

ab.12.cd   or   e.12.fg

where a ~ g represent digits. We can see that the first combination
can't be a palindrome. The latest that the second combination could be
is December 9 (specifically, to be a palindrome 9.12.19), but that is
more than ten days away from the next January. Thus, the two dates
have to be in the same year.

If the dates are in the same year, then the last two digits of the two
dates (the year) are the same. Since both dates are palindromes, that
means that the first two digits have to be the same as well. The first
two digits could represent DD (a two-digit day), D.M (a single-digit
day and a single-digit month; January ~ September), or D.1M (a
single-digit day and a two-digit month, where the second overall digit
is the tens digit of the month; October ~ December). I'll assume that,
as in your examples, we won't represent single-digit days or months as
01, 02, etc. - though I don't think that it would change the result.

Now we'll break it into cases.

Case 1: DD -> DD
(Going from a date where the first two digits represent the day to
another date where the first two digits represent the day). Since the
two DD's have to be the same (the mirror image of the year), the best
we could do here is two dates a month apart. For example, 19.2.91 and
19.3.91.

Case 2: DD -> D.M
(Going from a date where the first two digits represent the day to a
date where the first two digits represent the day and month). The two
digits have to be the same, so we have to check things like:

31.ab.13   and   3.1.13
30.ab.03   and   3.0.03
:               :

and so on through the 20's and 10's (for the first day). 31 is out
because the month (ab) would have to be the month before January, and
we've already eliminated changing years. 30 is out because there's no
"month 0" for the second date. Test 29, 28, etc. and see what you get.

Case 3: DD -> D.1M
(Going from a date where the first two digits represent the day to a
date where the first two digits represent the day and tens digit of a
month October ~ December). Since the second digit in this case is a 1,
we have to test three values:

31.ab.13   and   3.1c.13
21.ab.12   and   2.1c.12
11.ab.11   and   1.1c.11

We can quickly see that the second and third combinations are more
than ten days apart, no matter what months we use (February is not an
option here). For the first combination, 31.ab.13 and 3.1c.13, the
second date's month has to be 10, 11 or 12. If the second date's month
is 10 or 12, then the first date's month has to be 9 or 11 (to stay
within ten days they have to be consecutive months), but neither
September nor November has 31 days. If the second date's month is 11,
then the first date's month has to be 10, but 31.10.13 is not a
palindrome.

Case 4: D.M -> DD
(Going from a date where the first two digits represent the day and
month to a date where the first two digits represent the day). Clearly
these two have to be in the same month. Since the first digit of the
second date is restricted to 1, 2, or 3, we have to check:

1.a.a1   and   1a.a.a1
2.a.a2   and   2a.a.a2
3.a.a3   and   3a.a.a3

Again we can quickly see that the second and third combinations are
necessarily more than 10 days apart. Since the digit a can't be 0
because the month has to be a), the best we can do is 1.1.11 and
11.1.11 - which you already found.

Case 5: D.M -> D.M
(Going from a date where the first two digits represent the day and
month to a date where the first two digits represent the day and
month). Since D and M must be the same for both dates (and by
necessity the year), this case can't generate distinct dates.

Case 6: D.M -> D.1M
(Going from a date where the first two digits represent the day and
month to a date where the first two digits represent the day and tens
digit of a month October ~ December). Here we have to examine:

a.1.1a   and   a.1b.1a

Clearly the closest these two dates could be is nine months apart.

Additionally, we have to consider the cases:

Case 7: D.1M -> DD     (similar to case 4)
Case 8: D.1M -> D.M    (impossible in the same year)
Case 9: D.1M -> D.1M   (similar to case 1)

I'll leave it to you to consider these cases and see what you get.

Incidentally, if you use the U.S. custom of MM.DD.YY, there are at
least three runs of nine consecutive palindromic days; October 1-9,
1901; November 1-9, 1911; and December 1-9, 1921. There may be more,
or perhaps longer runs - I didn't do an exhaustive check.

I hope this helps - if you have any more questions, write back.

- Doctor TWE, The Math Forum
http://mathforum.org/dr.math/
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