Smallest N Value for 4-Digit Consecutive Integers Divisible by 2010^2

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Homework Help Overview

The problem involves finding the smallest value of N such that the product of N consecutive 4-digit integers is divisible by 2010 squared. The integers must be within the range of 1000 to 9999, and the possible answers for N range from 4 to 12.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the prime factorization of 2010 squared and its implications for the consecutive integers. There is an exploration of the divisibility of specific integers and the constraints of the problem, particularly regarding the need for consecutive integers and their upper limit.

Discussion Status

The discussion is active, with participants offering hints and questioning assumptions about the number of consecutive integers needed. There is a focus on the factors of 2010 squared and how they relate to the integers being considered. Some guidance has been provided regarding the factors and their distribution among the integers.

Contextual Notes

Participants note that the integers must be consecutive and less than 10,000, which adds complexity to the problem. The requirement for multiple occurrences of the prime factor 67 is also highlighted as a critical aspect of the solution.

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

Hint: the tricky one is 67. :wink:
 
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?
 
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?
 
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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