On Aug 15, 2010, at 11:51 PM, Ian Kelly wrote:
On Sun, Aug 15, 2010 at 4:36 PM, Baba <raoul...@gmail.com> wrote:
Hi Mel,
indeed i thought of generalising the theorem as follows:
If it is possible to buy n, n+1,…, n+(x-1) sets of McNuggets, for
some
x, then it is possible to buy any number of McNuggets >= x, given
that
McNuggets come in x, y and z packs.
so with diophantine_nuggets(7,10,21) i would need 7 passes
result:53
but with (10,20,30) and 10 passes i get no result
You're on the right track. In the case of (10,20,30) there is no
largest exactly purchasable quantity. Any quantity that does not end
with a 0 will not be exactly purchasable.
I suspect that there exists a largest unpurchasable quantity iff at
least two of the pack quantities are relatively prime, but I have made
no attempt to prove this.
That for sure is not correct; packs of 2, 4 and 7 do have a largest
unpurchasable quantity.
I'm pretty sure that if there's no common divisor for all three (or
more) packages (except one), there is a largest unpurchasable
quantity. That is: ∀ i>1: ¬(i|a) ∨ ¬(i|b) ∨ ¬(i|c), where ¬(x|
y) means "x is no divider of y"
Cheers, Roald
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