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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?

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    hint: you multiply $x$ consecutive integers so that one...2012-09-01
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    There are no "concepts", just original thought. Try to prove that one of the brackets is divisible by $x$ by considering what $a$ is mod $x$.2012-09-01
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    Hint: can you think of any reason why $x$ must divide one of the numbers $a,\ldots,a+x$?2012-09-01
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    So I can use factorial notation?2012-09-01
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    so that one of them must be a multiple of $x$ (every $x$-th integer is a multiple of $x$).2012-09-01
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    Well, you get (at nearly but not quite the same price) that your products are all congruent to zero modulo $x!$, so you might as well prove that!2012-09-01

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