It seems to be one of those puzzles that can't be solved algebraically...well, maybe, but others will have to figure that out... Like Fermat's Last Theorem....AldenG wrote:That one wasn't too hard because there aren't that many numbers divisible by 11. Plus intuition tells you it has to be within a certain range because of how sums and products grow differently. And as you look at likely options there's a quick convergence toward the solution.
I'd say be less formal and try accessing the non-scientist within you.
I figured out the answer through good old trial and error....
You can "approach" a solution this way:
(100x + 10y + z)/11=x + y + z
which reduces to:
100x + 10y + z - 11x - 11y - 11z = 0
89x - y - 10z = 0
You know that all digits must be between "0" and "9"
So right away you know that "x" must equal "1" ...any higher number would not be possible, nor would "0" be possible.
So now we have:
y + 10z = 89 [Note: The highest possible number on the left side of the equation is "99"...so "2" is immediately eliminated as a solution for "x". ]
and now we can "solve" for "z"...only "8" is possible..."7" and "9" are either too low or too high.
And therefore "y" can only be "9"....
And there's the answer... "198"
If there is a more "elegant" mathematical solution to this it is beyond me....
And it would be rather painful to have to go through this if the multiple were, say, "111"
