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31 Jul 11
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?
If I understand, it is six. 2003 = 1000+1000+3. And 1000*1000*3=3,000,000. Is it a coincidence that 2003 is prime?
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31 Jul 11
Originally posted by kes29
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?
128, 625, 1250
The fact that the last factor looks roundish is not quite a coincidence. The fact that the first one looks roundish to me, and others of my vocation, certainly is not.
Richard
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01 Aug 11
Originally posted by PalynkaPartial answer, just to be difficult:
Now do it for 2011.
I get four answers, that is an a, b, and c, a second a, b, and c, a third a, b and c, and a fourth a, b, and c, each with the product of the multiplication having 4 terminal zeroes.
This draws on something I see in shallow blue's results for 2003, that may be a misinterpretation on my part.
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01 Aug 11
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)
01 Aug 11
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.
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.