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I consider the numbers one at a time and keep a running track of the minimum that I need:
In order for r/10 to be an int, r needs at least one 2 and one 5. So now we have 2*5. In order for r/10 to be an int, r needs at least one 3 and one 5. We already have a 5, so this just adds a 3; now we have 2*3*5.
so 2*3*5 is the absolute minimum that must be in r's PF. You can confirm that his works; 2*3*5 = 30. 30 is a mult of both 10 and 15 and there is no smaller number that is a mult of both 10 and 15.
So it is enough to have just a single 5. Only in case of successive division (when the 5 is canceled off in the first step) do you need two 5s. Since it is not mentioned that you need to divide success
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