Given a 2011 digit number, divisible by 9.
Let x be the sum of its digits and y the sum of the digits of x.

Find z, sum of the digits of y .

It seems clear to me that the solution should be 9 since we are essentially finding the digital root.   The only thing to check is that there is no starting number that hasn't settled at 9 after three steps.

The smallest 2011 digit number divisible by 9 would be a 1 followed by 2009 0's and ending with 8.  x=9, y=9, z=9.

The largest 2001 digit number divisible by 9 would be all 9's.  x=2011*9=18099, y=27, z=9.
The key result is that 18099 is a largest possibility for x.

Can we construct a number where z is not 9?  The next largest possibility would be 18.  The smallest possibility for y would then be 99.  x would then need s.o.d. of 99, the smallest possibility would be 99,999,999,999 which is far greater than 18099.

One way to reword the problem to have z not necessarily be 9 is to change the first sentence to:
Given a 11,111,111,111 digit number divisible by 9.

Posted by Jer on 2012-02-21 12:12:24