10*8 + 2*7 + 1*6=100
What coins are these?
I don't have anything better than trial and error, but I'd set it up like this:
Let there be q quarters, d dimes, n nickels, and p pennies. Then:
100 (cents) = 25q + 10d + 5n + 1p
and
q + d + n + p = 21
Solve the second equation for p:
p = 21 - q - d - n
and insert this value of p into the first equation:
100 = 25q + 10d + 5n + (21 - q - d - n)
79 = 24q + 9d + 4n
I would start with the assumption there are 3 quarters (which almost certainly won't work), and work my way down. It could take a while without some more insight, which I'm afraid I can't offer you.
-Dan
I am not sure what coins are you using.
If you have a lot of coints it becomes the Frobenius Coin Problem.