5
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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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    Every natural number is divisible by $1^3$.2012-08-25
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    i think that as @ Chris Eagle said only solution is $1$,otherwise you have to know that,among these $2012$ consecutive numbers,some of them is prime,which can't be divides by any number or even by it's cube2012-08-25
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    @dato That's not what Chris said ("only" solution). Chris is getting at the $\ge1$ thing (which should be written $>1$ or else the problem is trivial). What makes you think a stretch of $2012$ consecutive numbers necessarily contains a prime?2012-08-25
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    @anon i understood what he said,just i have added about prime numbers2012-08-25
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    consecutive number i think means any number +1 right?2012-08-25
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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

1 Answers 1

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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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    How does that work with $x\equiv 0\bmod{3^3}$ and $x+3\equiv 0\bmod{9^3}$?2012-08-25
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    @SiliconCelery Where are you getting $9^3$ from? The moduli that Andre is using are the cubes of *prime numbers*, and $9$ is not prime. It would be $x+3\equiv0~\bmod 7^3$, since $7$ is the next prime number after $5$.2012-08-25
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    @SiliconCelery: Thanks, you pointed out a possible ambiguity in what I meant by and so on. It is the primes cubed. I changed things so that this is clearer.2012-08-25
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    I'd add that the fact that the numbers on the right hand side are cubes of distinct primes and therefore all co-prime means the Chinese remainder theorem will find a solution. Finding the smallest solution seems very difficult.2014-05-09
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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