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A fun and interesting math puzzle 
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#1
Feb1413, 10:17 PM

P: 2

Here's a nice challenge that I made up thinking about compressing data one night...
Print out a number that you can read within it: The numbers 1 to 20(easier) or 1 to 99(very hard) Rules: →Can be read forward or backwards and use same numbers eg. 123 you can read 1,2,3,11,12,22,21,23,32,33 →Cannot! have any two digit repeats eg. 121 you can read 12 foward and also backwards! exception! any double numbers like 11 22 33 etc. The smaller the answer in digits the better. (easy) Here is the answer I get going from 1 to 20 starting with writing a 0123456789 down. 1520123456789113416718 and again starting with writing 8642013579, 17418642013579112516 Is there a better number out there that is shorter yet? (hard) starting with the number 10 I get: (10 doesn't have to end up at the start) 795875564736032972691715211041361802248243358394450776688990 This is the best attempt so far I've gotten... I see the number 24 repeated right off the bat so this is wrong. This is turning out to be quite interesting and quite a fun challenge. I would love to know if there is only one unique answer when you start by writing the number 10 down. What would happen if you started with the number 61? I would love to see some answers (either right or wrong!). 


#2
Feb1513, 03:47 PM

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P: 11,617

It does not matter where you start, as every number has to occur.
Here is a shorter string for 1 to 20: 918716514311201 I can show that it is optimal: With one substring "11" and n other occurences of "1", we can form at most 2+2n twodigit numbers with a 1 inside. We need 9 of them (not counting 11), therefore n>=4. This means we need at least six times a "1", and one time all other digits of course. This gives at least 15 digits. Modifications: The 9 can be moved to the other side, you can move blocks of 23 digits from left to right, and all digits from 3 to 9 can be exchanged in their position in arbitrary ways. And you can mirror the whole string, of course. A similar string for 199 cannot exist. We just have two positions on the edges, this leaves at least 7 digits from 19 which are not there. There are 18 numbers containing those digits once, which gives 9 pairs (where one pair repeats the number). This is not possible, the number of pairs has to be even. 


#3
Feb1613, 02:59 AM

P: 2

That sure saves me racking my brains from trying get that last 99 in there! Thanks!



#4
Feb1613, 12:04 PM

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P: 11,617

A fun and interesting math puzzle
Here is another puzzle: We already know that it is not possible for 199 (or any number larger than 99), and we know that it is possible for 120. What is the largest number where such a string is possible?



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