I used to be stuck for hours a day in traffic jams. To fight
boredom, I started playing with the numbers on the license plates of the
cars around me. In my defense, let me say that I didn't invent this
unconventional pastime. Ramanujan was busy at it long before me.
The license plates in the area where my rush-hour traffic jams were happen to contain no letters. They contain only a string of seven digits. My particular game was to try and represent this string as a concatenation of n<5 odd-length digit palindromes. A palindrome is a string that can be read backwards to yield the same string as when being read forwards. As an example, "able was I ere I saw elba" is a letter palindrome. An odd-length digit palindrome is a palindromic number which is composed of an odd number of digits in decimal notation. As you can see, there is no point in creating a 7-digit string as a concatenation of n≥5 odd-length palindromes, as most of the digits end up being part of 1-digit palindromes. 1-digit palindromes are still allowed, but I tried not to overuse them... In any case, trying out this game, I found out, at first, that only for a very small percentage of the cars was there a winning solution. I therefore added a small "cheat" to the rules. I allowed myself to change one digit to a digit of my choice, before attempting to break down the original number into a concatenation form. The question this month is this: how many different seven-digit strings exist, which, with the change of at most one digit, can be represented as the concatenation of less than 5, odd-length, palindromic numbers? Because this question can easily be answered by a computer, there's a twist: answers will be accepted only if they satisfy the following conditions.
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List of solvers:Oded Margalit (2 November 18:00)Omer Angel (4 November 16:00) Itsik Horovitz (8 November 20:00) |
Elegant solutions can be submitted to the puzzlemaster at riddlesbrand.scso.com. Names of solvers will be posted on this page. Notify if you don't want your name to be mentioned.
The solution will be published at the end of the month.
Enjoy!