The Question ( http://www.ecst.csuchico.edu/~kend/potw/index.html )

1.Take a 2-digit number and multiply the two digits together. Repeat with each product until a single digit is the result. For most starting numbers, no more than three of these steps are needed to arrive at the final single digit. (For example: 57: 5*7=35: 3*5=15: 1*5=5). Find one that takes four steps.

The ONLY two digit number that takes four steps is 77.
7*7=49; 4*9=36; 3*6=18; 1*8=8

Believe me, it was only the fifth number I tried. However, it inspired me to write a small tcl program to find such numbers with higher number of digits.

Let us denote the number of steps it takes to reduce a n digit number to a single digit as "hardness"
Conjecture: Any n digit number has a hardness less or equal to  n+2.

The above conjecture holds good for n<=6 (that was as far as I could push tcl).

 Number of digits (n) hardness Digit sets* 2 4 {7,7} 3 5 {6,7,9}  {6,8,8} 4 6 {6,7,8,8} 5 7 {6,8,8,8,9} 6 8 NONE

Some interesting findings:

As seen in the above table, There are two sets of three digit numbers which have a maximum hardness of 5
Surprisingly, there was no six digit number with a hardness of 8

 Clockwise from top left: a) Graph showing the hardness of numbers between 10 and 99, b) numbers between 100 & 999 and c) numbers between 1000 & 9999. Note that the max hardness numbers are clustered around 6,7,8,9 in  each level. Anyone see any other pattern?

2.Start with a 3-digit number. Choose any one of the digits and remove it to make a 2-digit number. Multiply the 2-digit number by the digit you removed. Eventually, you will get a 2-digit number, which will yield a single digit, as above. What number should you start with, and what steps should you take to find the longest series of steps you can take to arrive at that final single digit?