Question

Consider this string of digits:
A=03161011511417191111
It has two, twelve 1's, zero 2's and so on.
We construct another string of digits, called B, as follows: write the number of 0's in A, followed by the number of 1's, followed by the number of 2's, and so on until we write the number of 9's. 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.

a) Find all the strings that have 1000110001 as their derived string.
b) Show that if a string has less that 1000 digits, then its derived string has at most 29 digits.
c) Prove that if the starting string has only two digits, then the sequence of strings will eventually repeat.
d) construct the sequence pf derived strings that start with the string 77377573677407172087777

Answer 1

a) just write all the combinations with the digit 0,4,5, and 9