Math "non loop" equation

Question

I am creating a Java program but am dealing with a math problem but can't seem to solve it. This is my assignment:

I have infinite numbers of numbered papers(1,2,3, ..). 
This papers are stacked into stacks using 2 variables a and b.

If a = 5 and b = 3 the stacks look like:

So the first stack has "a" items (5) and each next has "b" more (3) as previous.

Now I have to figure out an equation that would tell me how many papers I need to remove to get to the specific one.

For example, if I want to get to the paper numbered with 20, I will need to remove 4 papers to get to it. And this is the solution I am looking for, when I would enter a number of a specific paper, I just need to know how many I need to move (in the stack that it is in) to get to it. When I have this I put all papers back and search for the next one.

Since I will make a program with it, that will deal with long numbers (up to 16 digits), the best solution would be to find some "non loop" equation. For example, if I will set a = 34354 and b = 56774 and I want to know how many papers to remove to get to paper numbered with 533663634611112.

My inputs will be:
- numbers a and b
- number of papers I want to reach (c)
- regarding the above variable (c), I will input this many numbers that represent numbered papers

Output:
- number of papers I have to move to get to all the papers that I have inserted

All number will be very long, so would be the best to make a non loop program. Thanks for all help about this issue.

Answer 1

Mathy answer, but it might be enough to help you:

Stack number is the single integer solution of two inequalities:

ax + b(x-1)> argument AND a(x-1) + b(x-2) < argument

Whether that can be computed without using a loop I cannot say and I don't know of any library that could help you (though I am not an expert in math Java libraries). After that the answer you're looking for is

a*x + b(x-1) - argument

Answer 2

You don't need any loops for this. The number of papers in the nth pile is a + (n-1)b. So the number of papers in the first n piles is (arithmetic series) (b/2)n^2 + (a - b/2)n.

Using this, find the number of piles needed to get your paper. In your example, set this equal to 20 (and use 3 and 5). So you have 20 = 1.5n^2 + 3.5n. Solving this (using the quadratic formula) gives n = -5 and n = 8/3. Ignore the negative one as it doesn't make sense in this situation.

If you need 8/3 piles to get to your paper, it must be in the 3rd pile (round up to nearest int (ceiling function)). Use the formula above with n = 3 to find there are 24 papers in the first three piles (1.5 * 3^2 + 3.5 * 3 = 13.5 + 10.5 = 24).

24 - 20 = 4 = the number of papers you need to remove.

mockup solution, you need to remove hardcoding and make sure nothing overflows for the large numbers

Answer 3

You need some sort of loop to do this. I can help with the math, but someone else may have to help with the programming

j=1
TOTAL = ja + (j-1)b
If the TOTAL is less than c, then let j=j+1 and loop
If the TOTAL is greater than c, end the loop (this loop shouldn't take too long)
Then calc TOTAL - c from it to get your answer

That method should work. But a presume a loop like this shouldnt take much power even with long numbers.