Product of all the factors of a 129600 divisible by 5
Answers
129600=26×34×52129600=26×34×52
The factors of 129600129600 that are divisible by 55 either have 5151 or 5252 in their prime factorisation.
So we find all the permutations of 26×3426×34, then multiply each of these by 5151, and then by 5252. Then we multiply everything out.
The number of possibilities of 26×3426×34 are:
30×20,30×21,…,30×2630×20,30×21,…,30×26;
31×20,31×21,…,31×2631×20,31×21,…,31×26;
⋮⋮⋮⋮⋮⋮⋮⋮
34×20,34×21,…,34×2634×20,34×21,…,34×26.
Multiplying these out, we have:
25(1+⋯+6)×37(1+⋯+4)=2105×37025(1+⋯+6)×37(1+⋯+4)=2105×370.
There are 35 pairs of products above. So there are 35 factors having 5151 in their prime factorisation and another 35 having 5252 in their prime factorisation. So the final answer is
(2105×370×51×35)×(2105×370×52×35)(2105×370×51×35)×(2105×370×52×35)
=2210×3140×5105=2210×3140×5105
≈2.5417×10203≈2.5417×10203.