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Find the biggest positive integer $n$ such that $n$ is divisible by all positive integers smaller than the integer part of the cubic root of $n$.

I'm quite sure it's $420$, but I need proof for that.

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  • $\begingroup$ Do you mean divisible by all positive integers smaller than the integer part of the cubic root of $n$? $\endgroup$ – J.R. Feb 5 '14 at 20:14
  • $\begingroup$ Consider $840$. Or did you mean less than or equal to? $\endgroup$ – André Nicolas Feb 5 '14 at 20:19
  • $\begingroup$ Yes, indeed. My bad. $\endgroup$ – user2977079 Feb 5 '14 at 20:19
  • $\begingroup$ Less than. No equality case $\endgroup$ – user2977079 Feb 5 '14 at 20:19
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I think you mean "must be divisible by all positive integers less than its cube root". (Now confirmed in the comments.)

Suppose the number was at least 420. Then the floor of its cube root is at least 7, and hence it must be divisible by 1, 2, 3, ..., 7 and so must be a multiple of 420. If it is greater than 420 it must be at least 840 and hence its cube root is 8 and so it must be divisible by 840. But if it's at least 840 it must be divisible by lcm(1, 2, ..., 9) = 2520 and hence must be divisible by lcm(1, ..., 13) = 360360.

This pattern looks like it will continue, so let's prove it! If the next interval is at least twice as long as the last, it will contain a new prime (at least 11) and so the new interval will be at least 11 times longer than the previous. 11 > 8 so the cube root increases by a factor of > 2, and hence by Bertrand's postulate we get another prime next time. Done!

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