Closest Palindromic Dates

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.

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/