Question 693108
<pre>
10d + 5n < 100

  2d + n < 20

       n < 20 - 2d

And since we must have at least 1 nickel,

   0 < n < 20 - 2d

If there is 1 dime, d = 1, that inequality becoms

     0 < n < 18  and there are 17 ways to have that.

If there are 2 dimes, d = 2, that inequality becoms

     0 < n < 16  and there are 15 ways to have that.

If there are 3 dimes, d = 3, that inequality becoms

     0 < n < 14  and there are 13 ways to have that. 

And it goes down 2 each time we add a dime, and we
can have up to 9 dimes, so the answer is

17+15+13+11+9+7+6+5+3+2+1 = 81 ways.

Edwin</pre>