Only 1 output worng among all cases?

Case No. 5, input string is "43203 60 5", my output is "892", right answer is "899" I just want to make sure whether there is a mistake at your end. Kindly check, I've checked plenty of time, If not Kindly explain the case? Much appreciaited.

I've found out that there is a direct solution (i didn't get it myself, I just found out by searching, I solved the problem with while bucle like almost everybody).

Code

money = 10

cost = 2

extra = 2

boughtCandies = money / cost

result = boughtCandies + (boughtCandies - 1) / (extra - 1)

Does anyone know the math basis or where this comes from, or where can I read about this?

Thanks!

Very simple. 100% all test cases passed

   static int chocolateFeast(int n, int c, int m) {
        int wrapper=0;
        wrapper=n/c; 
        int choc_count=wrapper;
				while(wrapper>=m)
        {  
            wrapper=wrapper-m;//exchange
            wrapper++;//new choc wrapper..  so ++
            choc_count++;//increase choc count
        }
        return choc_count;
    }

The test case: 12 4 4 result is 3.
Why can't I borrow a Wrapper and when I finish mine, I return the Wrapper?

I came up with a nice formula for solving the problem.

We start off with a self-referential formula:

t = money + wrappers
t = floor(n/c) + floor((t-1)/m)

The reason we use t-1 over t is because the wrapper you acquire from your last exchanged chocolate is not part of your available resources.

We then solve for t:

t - floor((t - 1) / m) = floor(n/c)
ceil(t - (t - 1) / m) = floor(n/c)
ceil((mt - t + 1) / m) = floor(n/c)

We can convert this to an inequality. The idea is that mt - t + 1 must be at between floor(n/c)-1 (non-inclusive) and floor(n/c) (inclusive) multiples of m.

m(floor(n/c) - 1) < t(m - 1) + 1 <= m(floor(n/c))
m(floor(n/c) - 1) - 1 < t(m - 1) <= m(floor(n/c)) - 1
(m(floor(n/c) - 1) - 1) / (m - 1) < t <= (m(floor(n/c)) - 1 ) / (m - 1)

Note there can be multiple values of t that are within this range. We select the largest, because we want the maximum amount of chocolates

t = floor((m * floor(n/c) - 1) / (m - 1))

Another form of this is what @ansonete described below:

t = floor((m * floor(n/c) - floor(n/c) + floor(n/c) - 1) / (m - 1))
= floor(floor(n/c) + (floor(n/c) - 1) / (m - 1))
= floor(n/c) + (floor(n/c) - 1) / (m - 1)