Coupons equation

[-Demosthenes-]

Is there an equation for this problem?

 

Each candy bar comes with a coupon, if you get 7 coupons it's good for a free candy bar. What is an eqation that gives you the final amount of candy bars you get (including free ones) using X as the originial number you bought?

 

EX if I bought 27 candy bars, I'd get 27 coupons > 3 free bars + 6 left over coupons. But then those 3 will come with 3 more coupons, making the total amount of coupons 9, which is enought to get another candy bar. So, in the end I'd have 31 candy bars.

 

And it gets even more difficult with higher numbers. This question was given in class (too late to get any credit for it ), but I could never solve it.

[JesuBungle]

Errr. You had me workin on this for 3 hours and all I got was an indentation from a pencap on my forehead:-(

[-Demosthenes-]

I have no idea.

[uncool]

I believe the answer should eventually come out to

floor(x + x/b + x/b^2 + x/b^3) = floor(x*b/(b-1))

where x is the original number you had and b is the number you need to get a free bar.

In your example, it's floor(27*7/6) = floor(63/2) = 31.

=Uncool=

[uncool]

A reason for my answer below:

Let us define a function f(a, b) = floor(a/b) + f(a - b*floor(a/b) + floor(a/b), b)

f(a, b) = 0 if 0 <= a < b

I say f(a, b) = floor(a/(b-1)).

 

Clearly, all cases do go to the base case shown above. Now, we will use WOP to show that this is the case in general.

Let us assume that a is the lowest number such that f(a, b) =/= floor(a/(b-1)). Then f(a) = floor(a/b) + floor(a/(b-1) - b*floor(a/b)/(b-1) + floor(a/b)/(b-1)) = floor(a/b) + floor(a/(b-1) - floor(a/b)/(b-1) - floor(a/b) + floor(a/b)/(b-1)) = floor(a/b) + floor(a/(b-1) - floor(a/b)) = floor(a/b) - floor(a/b) + floor(a/(b-1)) = floor(a/(b-1)).

This function is equal to the number of bars you get if you start out with a coupons and redeem b coupons for a bar each time. Therefore, if you start out with ab bars, you get floor(ab/(b-1)) bars at the end.

=Uncool-