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Quite a triplet (Posted on 2015-07-02) Difficulty: 3 of 5
Let (n,n+1,n+2) represent a triplet of consecutive numbers, each having 5 distinct prime factors.

Find tne value of the lowest n.

No Solution Yet Submitted by Ady TZIDON    
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Silly Observation? | Comment 2 of 10 |
So to get at least 5 prime factors for each of (n, n+1, n+2), one  needs numbers >~10^6.  Also notice that for Charlie's program, there are no triplets of 6 prime factors each under 10^7.  I wonder if there is some type of trend here?  Meaning, <= q factors for some value of n.  I'm not saying this is necessarily related to powers of 10 at all, but I wonder what the curve looks like.  Also for (n,... n+m), where m<>3.

Edited on July 2, 2015, 6:51 pm
  Posted by Kenny M on 2015-07-02 18:49:15

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