$n$ divides a product of $n$ consecutive integers (by $n$ divides one of them) If $x$ is positive integer, prove that for all integers $a$, $(a+1)(a+2)\cdots(a+x)$ is congruent to $0\!\!\!\mod x$.
Any hints? What are the useful concepts that may help me solve this problem?
 A: Note that there is some $k$ such that $0\leq k< x$ and $a\equiv k\mod x$. Then
$$a+(x-k)\equiv k+x-k\equiv x\equiv 0\mod x$$
and what do you get when you multiply this by other stuff?
A: One of the numbers $a+1, a+2,\ldots, a+x$ is congruent to $0$ mod $x$.  Multiplying by $0$ yields $0$.
A: Hint $\, $ Any sequence of $\,n\,$ consecutive naturals has an element divisible by $\,n\,$. This has a simple proof by induction:  shifting such a sequence  by one does not change its set of remainders mod $\,n,\,$ since it effectively replaces the old least element $\:\color{#C00}a\:$ by the new greatest element $\:\color{#C00}{a\!+\!n}$
$$\begin{array}{}& \color{#C00}a,\, &\!\!\!\! a\!+\!1,\, &\!\!\!\! a\!+\!2, &\!\!\!\! \cdots,\, &\!\!\!\! a\!+\!n\!-\!1  & \\
                  \to & &\!\!\!\! a\!+\!1,&\!\!\!\! a\!+\!2, &\!\!\!\! \cdots, &\!\!\!\! a\!+\!n\!-\!1,\, &\!\!\!\! \color{#C00}{a\!+\!n} \end{array}\qquad$$
Since $\: \color{#C00}{a\,\equiv\, a\!+\!n}\pmod n,\:$ the shift does not change the remainders in the sequence. Thus the remainders are the same as the base case $\ 0,1,2,\ldots,n-1\, =\, $ all $ $ possible remainders $\!\bmod n.\,$ Therefore the sequence has an element with remainder $\,0,\,$ i.e. an element divisible by $\,n.$
More generally if the step-size $\,b\,$ is coprime to the modulus $\,n\,$ then the  arithmetic progression $\,a+k\:\!b,\ k = 0,\ldots,n\!-\!1\,$ is a complete residue system mod $\,n\,$ (so it contains a residue $\equiv 0,\,$ i.e. divisible by $\,n)$.
A: This has been already nicely answered, but here is another way to state an approach.
You will also be able to  notice that you do not need the last term $(a + x)$ to have the relation you want. 
What you hope to find is one of the factors divisible by $x$. How do do know you can find one?
Any of the factors (mod $x$) will be the equivalent of the remainder of $a$ when divided by $x$ plus the remainder of the added term when divided by $x$. 
The remainder of $a$, call it r, will be $1\leq r< x$. And each of the added terms (other than $x$) will be equal to itself, that is one of the numbers $1$ thru $x - 1$. 
So one of these sums (factors) will definitely exactly equal $x$.
