Prove that if the starting string has only two digits, then the sequence will eventually repeat. Include a convincing argument. Consider this string of digits: A=03161011511417191111 It has two 0s, twelve 1s, zero 2s, and so on. We construct another string of digits, called B, as follows: write the number of zeros in A, followed by the number of 1s, followed by the number of 2s, and so on until we write the number of 9s. Thus B=21201111101 String B is called the derived string of A. We now repeat this procedure on B to get its derived string C, then get the derived string of C, and so on to produce a sequence of derived strings. A=03161011511417191111 B=21201111101 C=2720000000 D=7020000100 E=7110000100 F=6300000100 G=7101001000 H=6300000100 Notice that the last string equals a previous string so the sequence of derived strings will now repeat.

Expert Answers

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if the starting string has 2 digits, it is either 2 times the same digit or 2 different digits.

 

In this case there will be either  1 2 in the sequence or 2 1

If there is 1 2, then the third string will be 0010000

Starting to the fourth one, the sequences will be 0100000000 (1 1 and 0 other digits) all the time therefore it repeats....

If there are 2 1, then the third sequence will be 0200000000

then the fourth :0010000000

Starting to the fifth, the sequences will  always be  0100000000 therefore it repeats.

 

In conclusion, the sequences repeats after the 5th sequence for any 2 digits initial sequence.

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