Question

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 0s 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 process 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=031611011511417191111
B=21201111101
C=2720000000
7020000100
7110000100
6300000100
71

Answer 1

7101001000
6300000100
Notice that the last string equals a previous string so the sequence of derived will now repeat.

Answer 2

Show that if a string has less than 1000 digits then its string has at most 29 digits.

Answer 3

the above ans is correct as well as this one
7101001000
6300000100
Notice that the last string equals a previous string so the sequence of derived will now repeat.
Show that if a string has less than 1000 digits then its string has at most 29 digits.

Answer 4

So therefore we must prove it cannot have 30 digits or above

Answer 5

The derived string is made up of 10 blocks of numbers.
[# zeros] [# ones] [# twos] [# threes] ...
If all of the numbers were the same in the 999 digit input, then one of the blocks would have length of three. The rest would be one, for a total length of 3 + 1 * 9 = 12. What if it's half and half? 3 + 3 + 1*8 = 14. Continue, and notice that it's most efficient when we split it up. Why?

Well, think of it as a game where the object is to get another digit in a block. When we first start out, the cheapest option is to get to 10 in total. 10 gets us two digits in a block. If we want another digit in that block we must spend 90 more numbers instead of 10 somewhere else! So, we distribute it equally until we need to work on getting hundreds.
999 digits
-10 for 0
-10 for 1
-10 for 2
etc...
-100 in total so we have 899 digits left to use. Our length so far? 2 * 10 = 20.

- 90 for 0
- 90 for 1
...
- 90 for 8
- 90*9 = 810 in total (0 to 8) so we have 899 - 810 = 89 digits left. How long is our string now?

Well, it started out with a length of 20. Then, we got NINE numbers to a length of three. So, our length is now 29. With only 89 digits left, we don't have enough to get the final block to 3 digits long.Therefore, the derived string of a number less than 1000 can be no longer than 29 digits.

Answer 6

I really liked this problem. Where'd you get it?

Answer 7

it was year 8 maths homework=D

Answer 8

mwahahaha even though that was my exact question for my year 7 maths challenge but I do not get anything about what bluepig said ;-0

Answer 9

I don't get this :(