Originally posted by JS357
how kes knew it was right.
If you take the values of a, b and c such that they were as "close" together as possible (so that their product was maximized), e.g., a = b = 668, c = 667, you'll get a * b * c = 297631408 < 10^9. So the answer must be less than 9. And since the 3 numbers provided by Shallow Blue sum up to 2003, and their product has 8 terminal zeroes, it is obvious 8 is the correct answer.
How to find the numbers? Actually, I don't have the solution. I borrowed this problem from a Math Olympiad for the 17-year olds, which took place a few years ago. I think they had to solve it in about 20 minutes, so it is not supposed to be very difficult (admittedly, the kids are terribly smart nowadays!).
Perhaps, one might think as follows: OK, we want as many powers of 2's and 5's as possible. Now 2003 is odd, so we must have one number odd, and the other two even. How about the odd one? Obviously, it must end in 5. Maybe a power of 5? 5^2 = 25 - too small, 5 ^ 6 = 15625 - too large, 5 ^ 4 = 625 - looks OK... What about the even numbers? It is reasonable to assume, that one of them might end in 0, then the other one should end in 8. Again, 2^3 =8 - too small, 2 ^ 11 = 2048 - too large, 2 ^ 7 = 128 - looks fine. And then the third number can be found. Probably one needs to play with the calculator a bit, I'm not aware of any technique that could be used to solve problems of this nature.