What is the smallest smallest positive integer $n$ such that $n$ is divisible by $1$ to $100$ and $n$ is not $0$?


closed as off-topic by Rand al'Thor, JonMark Perry, Chowzen, w l, rhsquared Oct 29 '18 at 13:50

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  • $\begingroup$ I suggest write "smallest positive integer" instead, since "whole numbers" can sometimes include all integers (actually, integer means whole in Latin), and there's no smallest integer (even without zero). $\endgroup$ – trolley813 Oct 29 '18 at 9:34

Well, if you accept:

0 as a natural number, that's it.


You want lcm(1,2,...,100), which is the product of the maximal prime powers less than 100. This is 64*81*25*49*11*13*17*19*23*29*31*37*41*43*47*53*59*61*67*71*73*79*83*89*97 which is 69720375229712477164533808935312303556800.

  • 1
    $\begingroup$ 40 seconds ahead, so upvoting $\endgroup$ – trolley813 Oct 29 '18 at 9:25
  • $\begingroup$ And still another one before me! Why am I so blind to the obvious??? sorry +1 too for you $\endgroup$ – Omega Krypton Oct 29 '18 at 15:06

It depends on the definition of natural number. If zero is included into natural number, it's the obvious answer (because 0 is the smallest natural number at all, and is divisible by all other natural numbers). If not (so, natural numbers start from 1), so the answer is LCM(1,2,...,100)=2^6*3^4*5^2*7^2*11*13*17*19*23*29*31*37*41*43*47*53*59*61*67*71*73*79*83*89*97=69720375229712477164533808935312303556800 (this is the product of all primes below 100 in the maximum powers, which don't exceed 100)

  • $\begingroup$ Unlucky, better luck next time! :D Have a consolation upvote. (I'm lucky I managed to type out and recall my primes below 100 quick enough!) $\endgroup$ – boboquack Oct 29 '18 at 9:27
  • $\begingroup$ @boboquack Thank you. I've probably spent too much time on making a LaTeX-like formula from it, but unfortunately they (at least, such long ones) make spoilers' content visible. $\endgroup$ – trolley813 Oct 29 '18 at 9:31
  • $\begingroup$ Just realised that there's an answer so similar to mine posted before me, sorry! +1 $\endgroup$ – Omega Krypton Oct 29 '18 at 15:05

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