Nonnegative Integer Solutions for a+2b+4c=10^30 - Homework Help and Explanation

[saubbie]

Homework Statement

I need to find the number of nonnegative integer solutions to the equation a+2b+4c=10^30

Homework Equations

The Attempt at a Solution

I was thinking of trying to find perhaps some sort of a multinomial generating function, but am not sure how that will help me. Any suggestions? Thanks.

 

[jhicks]

Here are my thoughts:

You can find the total number of solutions by incrementally varying parameters. Since for any solution c will be the least likely to be an integer if you choose arbitrary a and b, varying c and trying to peg possible a's and b's would be the wisest way to proceed. For example:

If c=50, a+2b = 10^30-200. Because once again b is least likely to be an integer if we choose arbitrary a subject to a+2b=10^30-200, b would be the best to vary. Anywhere from b=(10^30-200)/2 to b=0 will have a corresponding a, so there are (10^30-200)/2+1 different solutions for the case c=50. Can you see how this result might generalize to all c?

 

[saubbie]

So for all c, there would be (10^30-4c)/2+1 corresponding to every possible c value, but that would give an infinite number of solutions wouldn't it?

 

[jhicks]

If 10^30-4c < 0, do you have any solutions for non-negative b and a? My approach is simply iteratively inspecting the sum and assuming a varying "chunk" of 10^30 is made up of 4c. The answer is definitely finite.