There are only 5 prime integer factors.

There are 44 integer factors. These guys:

1 1 1 1 358358
1 1 1 11 32578
1 1 1 13 27566
1 1 1 14 25597
1 1 1 143 2506
1 1 1 154 2327
1 1 1 179 2002
1 1 1 182 1969
1 1 1 2 179179
1 1 1 22 16289
1 1 1 26 13783
1 1 1 286 1253
1 1 1 358 1001
1 1 1 7 51194
1 1 1 77 4654
1 1 1 91 3938
1 1 11 13 2506
1 1 11 14 2327
1 1 11 179 182
1 1 13 154 179
1 1 2 11 16289
1 1 2 13 13783
1 1 2 143 1253
1 1 2 179 1001
1 1 2 7 25597
1 1 2 77 2327
1 1 2 91 1969
1 1 7 11 4654
1 1 7 13 3938
1 1 7 143 358
1 1 7 179 286
1 1 7 22 2327
1 1 7 26 1969
1 11 13 14 179
1 2 11 13 1253
1 2 11 91 179
1 2 13 77 179
1 2 7 11 2327
1 2 7 13 1969
1 2 7 143 179
1 7 11 13 358
1 7 11 26 179
1 7 13 22 179
2 7 11 13 179

Each of those examples has the product 358358 and each of those
examples is composed of integers. With one exception, they are not all
prime numbers, but remember that Skip asked for:

> 358358=*/x1, x2, x3, x4, x5
> How many different combinations of 5 integersx1, x2, x3, x4, x5  solve this
> equality? (Different orders don't count)

And it seems to me that there are 44 of those?

Thanks,

-- 
Raul

On Fri, Nov 3, 2017 at 7:25 AM, Erling Hellenäs
<[email protected]> wrote:
> Hi all!
>
> Raul:
>
> "Hmm... actually, thinking about it, the par approach here is not
> efficient enough for this example. 5 parRuskeyE 32 is too big of a
> result, I think. (358358 has 5 distinct prime factors and, thus, 32
> integer factors.)"
>
> However there are only 5 integer factors:
>
>    q: 358358
> 2 7 11 13 179
>    */q:358358
> 358358
>
> Now you ask me for 44 5 integer factorizations.
>
> As far as I understand there is only one 5 integer factorization of 358358
> unless you count 1 and 358358 as factors.
> In any case it can be handled by parRuskeyE.
>
> Cheers,
>
> Erling Hellenäs
>
>
>
> Den 2017-11-03 kl. 10:58, skrev Raul Miller:
>>
>> I'm not sure where you showed the 44 different 5 integer
>> factorizations of 358358?
>>
>> Thanks,
>>
>
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