1. thats fair enough but 176/8 was just an examlpe...

this needs to work for any number stord in the array up to (2^512) and your method woudnt work for large numbers. Also the output has to be read into an array such that each digit is stored in a spearte array element.

I dont think im explainig myself very well....

2. Again, you certainly shouldn't ask your teacher for help with this. There's no way they'd be of assistance.

3. its too late for that..

4. What you need to do is figure out the math first, THEN worry about the C++.

Draw these arrays on paper. On paper, can you perform the operations? If not, this is a mathematics issue and not a C++ issue.

If you understand how, mathematically, to break these numbers up and perform the operations on them, but you are having difficulty with C++ syntax, that is what this board is for. Of course, we will try to help where we can with math, but issues that are entirely mathematics oriented should be directed elsewhere.

5. Originally Posted by HU_RUN_TINGS
its too late for that..
Why would you start a program when it's too late to get help if you obviously have no idea what you're doing? Learn to prepare or you're going to have a hell of time passing the class.

6. Please read our Homework Policy. Please do not expect people to write the program for you. Show us what you have so far and people will help.

Please stop bumping this thread. Any further posts that are obvious attempts to bump (such as "....") will result in the closing of this thread and the deletion of all future threads related to this.

7. The OP might also benefit from reviewing such links as:
Cboard Posting FAQ
How To Ask Questions The Smart Way

Originally Posted by HU_RUN_TINGS
dividing (2^512)+1 by every number from 0...(2^512)+1
Well, I can guarantee that dividing by 0 will be bad, and dividing by (2^512)+1 will be 1. Actually, you only need to check divsors up to the square root of (2^512)+1.

8. To get you kinda pointed in the right direction, think about some of the math.

To obtain an integer result from x / y = z, both x and y would have to share a common denominator. The least of which would be 2. Continue finding a common denominator until y == 1 and therefore x == z.

Now all you have to do is pseudo code out the process of long division to handle the arrays, and you're all set. Of course it's a brute force method which may not be the most efficient way to go about it, but it'll move you in the right direction, and you can probably manage to get a little experience in recursion as well.

9. Slightly related, are there any good implementations of an int class without a size limit?

10. I use bignums
comes with free lcc compiler