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N Integer question

  1. Feb 25, 2010 #1
    1. The problem statement, all variables and given/known data
    The product of N 4-digit consecutive integers is divisible by 2010^2. What is the smallest N value? Multiple choice answers range from 4 to 12.


    2. Relevant equations

    N/A

    3. The attempt at a solution

    I tried multiplying the smallest combo possible 1000x1001x1002x1003 and quickly realize my calculator can't handle it. Looking for a head start on this question.
     
  2. jcsd
  3. Feb 25, 2010 #2

    tiny-tim

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    Hi Hockeystar! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    ok … head start … what are the prime factors of 20102 ? :wink:
     
  4. Feb 25, 2010 #3
    2,3,5,67. Square these number and you get 4,9,25,4489. I'm still stuck. Would the answer be 4 because there are 4 prime factors?
     
  5. Feb 25, 2010 #4

    tiny-tim

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    No, because they have to be consecutive (they also have to be less than 10,000).

    Hint: the tricky one is 67. :wink:
     
  6. Feb 25, 2010 #5
    1005 is the first integer with 67 as factor. Could 1005,1004,1003,1002,1001,1000 be the answer? 1005 is divisible by 67, 1004 by 4, 1002 by 3 and 1000 by 5?
     
  7. Feb 26, 2010 #6
    If you have a sequence of less than 68 consecutive integers, how many could be divisible by 67? Now how many times does 67 appear in the prime factorization of 20102?
     
  8. Feb 26, 2010 #7

    tiny-tim

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    Hi Hockeystar! :wink:

    Yes, Tedjn :smile: is right …

    if you start with 1005, all you've proved is that N ≤ 68, because you need two 67s in that sequence.

    So how should you start a sequence with N less than 68 ?​
     
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