## 18024Re: additive combinations all prime?

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• May 9, 2006
--- "Patrick Capelle" <patrick.capelle@...> wrote:
Something interesting with the product of the numbers ?
(1,4)--> 4 = 2^2
(1,4,8)--> 32 = 2^5
(3,5,8,13)--> 1560 = 2^3 * 3 * 5 * 13
(3,10,12,15,27)--> 145800 = 2^3 * 3^6 * 5^2
(5,30,33,42,60,63)--> 785862000 = 2^4 * 3^6 * 5^3 * 7^2 * 11

In each case the product of the n numbers gives a number whose
number of different prime factors (in the factorization) is smaller
or equal to n. Is it always the case ?
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--- "Patrick Capelle" <patrick.capelle@...> wrote:
It holds for n = 6 :
( 5,30,33,42,60,63)--> 785862000 = 2^4 * 3^6 * 5^3 * 7^2 * 11
(30,33,35,60,63,72)--> 9430344000 = 2^6 * 3^7 * 5^3 * 7^2 * 11
(30,33,47,60,72,75)--> 15075720000 = 2^6 * 3^6 * 5^4 * 11^ * 47
(30,42,47,60,63,75)--> 16788870000 = 2^4 * 3^6 * 5^4 * 7^2 * 47
(15,42,48,57,70,75)--> 9049320000 = 2^6 * 3^5 * 5^4 * 7^2 * 19
(30,33,42,60,75,77)--> 14407470000 = 2^4 * 3^5 * 5^4 * 7^2 * 11^2
(15,33,42,57,70,90)--> 7465689000 = 2^3 * 3^6 * 5^3 * 7^2 * 11 * 19
(15,35,42,57,72,90)--> 8144388000 = 2^5 * 3^7 * 5^3 * 7^2 * 19
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--- "Mark Underwood" <mark.underwood@...> wrote:
I assumed you meant it would hold for first cases. Afterall it
certainly doesn't hold for some cases after the first case. For
instance

14 + 9 = 23
14 - 9 = 5

And 14*9 has three different prime factors.

I'm going to try to look for solutions for n=7 and all 64 additive
combinations yielding primes. My computer is so slow however it does
not look promising.

kind regards,
Mark
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Important precision.
Thank you Mark.
I only started with the examples that you gave, without thinking to
other cases.
When we look at each n it is possible that it holds only for the
smallest set(s).
Or for all the sets when n is not too small ?

Best regards,
Patrick Capelle.
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