At least how many consecutive numbers should be taken?

Question

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 .

Answer 1

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.

Answer 2

I personally believe that you need only one number: $5040$, but I assume that is not the kind of answer you want.

Algebraically: $n = 5040$

The next best is four consecutive integers:
Let $x$ be the smallest integer:
$x(x+1)(x+2)(x+3)= 5040$
I would prefer graphing it but here goes the algebraic way:
$x^4 + 6 x^3 + 11 x^2 + 6 x = 5040$ Solving this gives: $x = 7, x= -10, x= \frac 32 i (\sqrt(31) +i), x = -\frac 32 i (\sqrt(31) - i)$
$x = 7$ is a valid answer as $7*8*9*10 = 5040$

$\therefore$ Four is the (second) least number of consecutive positive integers whose product is $5040$.

$n = 7$ is the least value iff (if and only if) $x = 1$.