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