How can I find a seven-digit number with all of its prime factors less than 20?

I have no clue how to do this.

  • 1
    $\begingroup$ It would depend on where one lost it.. $\endgroup$ – Daniel Lichtblau Feb 4 '16 at 22:48
  • 1
    $\begingroup$ I'm voting to close this question as off-topic because not specific to Mathematica software. $\endgroup$ – Daniel Lichtblau Feb 4 '16 at 22:49
  • $\begingroup$ @DanielLichtblau Let's suppose you didn't lose it. Then find it ad absurdum. $\endgroup$ – Dr. belisarius Feb 4 '16 at 22:55
  • $\begingroup$ In case someone visits this (badly asked) closed question, it seems to be about smooth numbers. A small code, yet too big for this marginal comment, shows there are 11447 seven-digit numbers composed of only primes less than 20. The smallest is 1000000, the largest is 9997020. $\endgroup$ – KennyColnago Mar 20 '16 at 1:14

Given that the 8th prime is less than 20 (Prime[8] = 19), then the product of all primes up to and including that one is:

Times @@ Prime[Range[8]]

(* 9699690 *)

You can list those primes:


You can answer the generalized problem from the below figure, e.g., How many distinct, sequential, smallest prime factors are needed to get a number greater than $10^{18}$?

ListLogPlot[Table[{i, Times @@ Prime[Range[i]]}, {i, 1, 20}],
 AxesLabel -> {Text[
    Style["number of\n sequential prime\n factors of \!\(\*
StyleBox[\"x\",\nFontSlant->\"Italic\"]\)", 18]], Text[Style["\!\(\*
StyleBox[\"x\",\nFontSlant->\"Italic\"]\)", 18]]},
 ImageSize -> 500]

enter image description here

  • $\begingroup$ This is a great answer since it even works for distinct primes. Of course, if the primes can repeat, 2^20 gets my vote. $\endgroup$ – barrycarter Feb 5 '16 at 1:35
  • $\begingroup$ Well sure... $2^{20}$. $\endgroup$ – David G. Stork Feb 5 '16 at 1:52

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