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/
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