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

[Hockeystar]

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.

 

[tiny-tim]

Hi Hockeystar!

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

ok … head start … what are the prime factors of 2010² ?

 

[Hockeystar]

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?

 

[tiny-tim]

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

Hint: the tricky one is 67.

 

[Hockeystar]

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?

 

[Tedjn]

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?

 

[tiny-tim]

Hi Hockeystar!

Yes, Tedjn 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 ?​