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More Referential Numbers (Posted on 2005-12-10) Difficulty: 4 of 5
Numbers like 21200, 3211000, and 6210001000 are self referential numbers, where the first digit is the number of zeros in the number, the second digit is the number of ones, etc. These were presented in the Self-Descriptor problems.

There are also pairs of numbers like {230100, 311100} which are cross referential numbers, where the first digit of one number is the number of zeros in the other number, second digit of one number is the number of ones in the other number, etc. By length, {230100, 311100} is the smallest such pair. Can you find longer cross referential numbers?

There are also cyclic referential numbers like the set {4110100, 3300100, 4102000}. 4110100 describes 4102000, 4102000 describes 3300100, and 3300100 describes 4110100. Are there any other cycles?

No Solution Yet Submitted by Brian Smith    
Rating: 4.0000 (2 votes)

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Hints/Tips One longer number and some thinking... | Comment 1 of 3

One point about these numbers is that the sum of the digits equal the number of digits in the number they reference. It isn't stated that they are necessarily the same length. 

If they don't need to be the same length, we can generate these pairs by beginning with some larger digit and always fill in the zeroes later.

e.g. {21011000, 43100}{31011000, 430100}{41011000, 4300100}{51011000, 43000100}{61011000, 430000100}. {6101010000, 530000100}...(and if we are allowed zeroes in digits representing numbers greater than 9, i.e. more than a ten-digit number,{910100001000, 830000000100})

If they do need to be the same length, we can just add one to the first digit of the longer string and add a zero in front of the 1 corresponding to that digit until the numbers have the same number of digits, hence {51011000, 43000100}

This should be harder to do with cycles, but let's try starting with the existing cycle:

{4110100, 3300100, 4102000}

{6110100, 3300001, 410200000}

{6110100, 330001000, 610200000}

{611x0100, 44000100, 51002000} this breaks down no matter how I approach it in trying for three numbers of the same length. I am beginning to suspect this may not have another solution with a triple cycle. Perhaps a quadruple cycle...

  Posted by Eric on 2005-12-10 19:34:33
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