Q:

At least how many consecutive numbers should be taken?

Problem Statement :
At least how many consecutive positive integers should be multiplied so that the product is divisible by $5040$ ? 
This problem should be solved using algebra.   
My attempt :
Let the multiplication of n consecutive numbers is divisible by 5040 . So according to question : 
x*(x+1)(x+2)$\ldots$(x+n-1)=5040*m
How can I find the value of n ? Please help me . 

A:

The question is phrased poorly and it is not clear what is intended.  I believe the following is relevant:
Claim:  the product of any $7$ consecutive natural numbers is divisible by $7!$ and no number smaller than $7$ has this property.
Proof:  It is clear that no number smaller than $7$ works as $6!$ is not divisible by $7$, let alone $7!$, and it is (visibly) the product of $6$ natural numbers.  
To see that $7$ has the desired property, let $m$ denote the start of the consecutive block and we remark that $$\binom {m+6}{7}=\frac {(m+6)(m+5)\cdots (m)}{7!}\in \mathbb N$$ is an integer, hence the desired result.

