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I found this task followed by a hint,
that I should try to apply Chinese remainder theorem to that:

Prove, that there exist 2012 consecutive natural numbers,
which satisfy that every one of them is divisible by a cube of a natural number $\ge$ 2.

The problem is, I don't really see how to use the theorem above. Can anyone help?

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    @dato yes but what anon is trying to convey here is that there doesn't have to be a prime every$2012$consecutive numbers.2012-08-25

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Look at the system of congruences $x\equiv 0 \bmod{2^3}$, $\,x+1\equiv 0\bmod{3^3}$, $\,x+2\equiv 0\bmod{5^3}$, $x+3\equiv 0\bmod{7^3}$, $\,x+4\equiv 0\bmod{11^3}$, and so on.

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    @gnasher729: You are right, the above construction based on CRT will undoubtedly produce a number $x$ enormously larger than the smallest possible number with the required property. Getting even basic size information about the smallest number seems pretty hopeless.2014-05-09