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?

  • 1
    $\begingroup$ Every natural number is divisible by $1^3$. $\endgroup$ – Chris Eagle Aug 25 '12 at 20:35
  • $\begingroup$ 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 cube $\endgroup$ – dato datuashvili Aug 25 '12 at 20:38
  • $\begingroup$ @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? $\endgroup$ – anon Aug 25 '12 at 20:39
  • $\begingroup$ @anon i understood what he said,just i have added about prime numbers $\endgroup$ – dato datuashvili Aug 25 '12 at 20:40
  • $\begingroup$ consecutive number i think means any number +1 right? $\endgroup$ – dato datuashvili Aug 25 '12 at 20:42

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.

  • $\begingroup$ How does that work with $x\equiv 0\bmod{3^3}$ and $x+3\equiv 0\bmod{9^3}$? $\endgroup$ – SiliconCelery Aug 25 '12 at 21:00
  • $\begingroup$ @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 $5$. $\endgroup$ – anon Aug 25 '12 at 21:07
  • $\begingroup$ @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. $\endgroup$ – André Nicolas Aug 25 '12 at 21:08
  • $\begingroup$ 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. $\endgroup$ – gnasher729 May 9 '14 at 15:31
  • $\begingroup$ @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. $\endgroup$ – André Nicolas May 9 '14 at 15:39

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