30 Jul '11 22:07>
Let a, b and c be positive integers such that a + b + c = 2003. What is the maximum number of zeroes that the product a * b * c can end?
Originally posted by kes29If the following is not hidden, don't read it. If you reply, don't reply and quote. only reply.
Let a, b and c be positive integers such that a + b + c = 2003. What is the maximum number of zeroes that the product a * b * c can end?
Originally posted by kes29Reveal Hidden Content
Let a, b and c be positive integers such that a + b + c = 2003. What is the maximum number of zeroes that the product a * b * c can end?
Originally posted by PalynkaPartial answer, just to be difficult:
Now do it for 2011.
Originally posted by iamatigerYes, it doesn't. But I like what I did anyway. "What is the maximum number of zeroes that the product a * b * c can end?" Four solutions, 4 zeros each. 2000, 10, 1; 1010, 1000, 1; 1760, 250, 1, 760, 1250, 1. And now I'm not telling how. 😛🙂
4 zeroes? That doesn't beat the obvious 1000 + 1000 + 11, which gets 6 (and I think is probably the best possible for this case)
Originally posted by JS357If 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 kes knew it was right.
Originally posted by iamatigerI get 8 with 2018!
Any Takers on 2015 or 2018?