Posers and Puzzles

Posers and Puzzles

  1. Joined
    26 Dec '09
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    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?
  2. Joined
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    31 Jul '11 04:58
    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?
    If the following is not hidden, don't read it. If you reply, don't reply and quote. only reply.





    Reveal Hidden Content
    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?
  3. Joined
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    31 Jul '11 12:38
    Your answer is wrong!

    It is a coincidence that 2003 is a prime, however the fact it ends with 3 - not entirely...
  4. Joined
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    31 Jul '11 15:09
    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?
    Reveal Hidden Content
    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
  5. Joined
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    31 Jul '11 20:51
    Yes, the numbers are correct (how did you get them, by the way?), and the answer is, of course, the number of zeroes that their product ends with.
  6. Joined
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    01 Aug '11 00:28
    Originally posted by kes29
    Yes, the numbers are correct (how did you get them, by the way?), and the answer is, of course, the number of zeroes that their product ends with.
    Inquiring minds want to know how S Blue got it and how kes knew it was right.
  7. Standard memberPalynka
    Upward Spiral
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    01 Aug '11 01:501 edit
    Originally posted by Shallow Blue
    [hidden]hidden[/hidden]

    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
    Now do it for 2011.
  8. Joined
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    01 Aug '11 18:29
    Originally posted by Palynka
    Now do it for 2011.
    Partial answer, just to be difficult:

    Reveal Hidden Content
    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.
  9. Joined
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    01 Aug '11 19:411 edit
    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)
  10. Joined
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    01 Aug '11 20:251 edit
    Any Takers on 2015 or 2018?
  11. Joined
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    01 Aug '11 20:44
    Originally posted by iamatiger
    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)
    Yes, 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. 😛🙂
  12. Joined
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    01 Aug '11 21:54
    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.
  13. Joined
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    01 Aug '11 22:06
    Originally posted by iamatiger
    Any Takers on 2015 or 2018?
    I get 8 with 2018!

    Reveal Hidden Content
    625 * 625 * 768 = 300000000
  14. Joined
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    01 Aug '11 22:44
    Kes, I don't think it is a coincidence that one of your numbers is the sum of two numbers that are both powers of 2.
  15. Joined
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    02 Aug '11 05:072 edits
    For bonus points:
    1) Give the first year that was a sum of 3 numbers that multiply together to end in 8 zeros.

    2) Give the first year that will be a sum of 3 numbers that multiply together to end in 9 zeros.
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