N Integer question

  • Thread starter Hockeystar
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  • #1
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Homework Statement


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.


Homework Equations



N/A

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.
 

Answers and Replies

  • #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:
 
  • #3
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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?
 
  • #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:
 
  • #5
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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?
 
  • #6
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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?
 
  • #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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