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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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re: more info | Comment 5 of 10 |
(In reply to more info by Ady TZIDON)

This still leaves open broll's question. Even when that site doesn't count repetitions, as in this current puzzle, subsequent numbers in the series can have repetitions even if they are not counted. When the number of consecutive numbers is 4 or more then of course the prime factor 2 must appear twice in one of their prime factorizations (i.e., be a multiple of 4).

In the current puzzle, if the beginning number is even, then either the beginning number or the ending number is a multiple of 4 and therefore has a prime (namely 2) that appears as a power (namely squared or higher).

The question then is, Can such a series of 3 numbers begin and end with odd numbers, and further, can broll's situation exist, where none of any of the prime factors involved with any of the numbers appear as a power higher than 1.

  Posted by Charlie on 2015-07-03 07:56:45
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