Take any whole number (larger numbers are more interesting)

1. x=the number of even digits, y=the number of odd digits, z=the total number digits.
2. Concatenate to create the number xyz (a leading zero is ok if x=0)

[Example: 12345678901 has x=5, y=6, z=11 for step 1 and 5611 for step 2]

3. Repeat steps 1 and 2 of this process until your number is unchanged.

What number(s) will the process terminate at?
Will the process always terminate?
Find a 20 digit starting number that terminates in the maximum number of steps (among 20 digit numbers.)

Any 20 digit number that has 10 even and 10 odd digits goes to 101020 --> 606 --> 303 --> 123 --> 123, so process ends after 5th iteration

Any 20 digit number that does not have 10 odd and 10 even digits terminates one step faster, as follows:
1st iteration: a 5 digit number ending in 20, because the even or odd digits in the original number (but not both) are less than 10
2nd iteration: a 3 digit number ending in 5
3rd iteration: 123 (because the the last digit of the previous number (5) is odd and so is excatly one of the 1st two digits)
4th iteration: 123 (process ends)

so the answer is :  Any 20 digit number that has 10 even and 10 odd digits

Posted by Steve Herman on 2006-01-02 18:25:08