Find $n$ consecutive, positive four-digit integers whose product is divisible by $2015^2$ . what is the least possible value of $n$?

Question

Suppose the product of $n$ consecutive, positive four-digit integers is divisible by $2015^2$. What is the least possible value of $n$?
$2015 = 5\times 13\times 31$
$65^2 = 4225$
I am not sure how to proceed. Thanks for help.

Answer 1

For the question in the title: You have three problems with the question. First, two consecutive integers have GCD of $1$, so can't both be divisible by the same number, including $2015^2$. Second, $2015^2 \gt 10^6$ so cannot divide any four digit number. Third, why ask for the minimum value of $n$? For a question like this you should be asking for the maximum value. In this case the maximum value is $0$.

For the question in the text: An upper bound is $14$. Take any four digit multiple of $961$. Find the multiple of $13$ below that and your list is from that multiple of $13$ to the next. The product of those $14$ numbers will be divisible by $31^2$ because one is, you have two multiples of $13$ and since the span is over $10$ you have two multiples of $5$. You just need to show that no four digit multiple of $31^2$ is close enough to a multiple of $13^2$ to make a shorter span. bof has shown that this fails because there is a shorter span.

Answer 2

To get $n=11$ consecutive $4$-digit numbers with product divisible by $2015^2=5^2\cdot13^2\cdot31^2,$ take the $11$ numbers from $2873=17\cdot13^2$ to $2883=3\cdot31^2.$

To see that $n=11$ is best possible, note that one of the numbers must be divisible by $31^2=961$ (or else you would need $n\ge32$ to catch two multiples of $31$) and one of them must be divisible by $13^2=169$ (or else you would need $n\ge14$). You can easily check that, among the $4$-digit multiples of $961,$ the one nearest to a multiple of $169$ is $3\cdot961=2883=17\cdot169+10.$