How many different ways can you make 82 cents using current u. s. currency

You can probably just work it out.

You need non-negative integer solutions to p+5n+10d+25q = 82.

If p = leftovers, then you simply need 5n + 10d + 25q ≤ 80.

So this is the same as n + 2d + 5q ≤ 16

So now you simply have to "crank out" the cases.

Case q=0 [ n + 2d ≤ 16 ]

Case (q=0, d=0) → n = 0 through 16 [17 possibilities]

Case (q=0, d=1) → n = 0 through 14 [15 possibilities]

...

Case (q=0, d=7) → n = 0 through 2 [3 possibilities]

Case (q=0, d=8) → n = 0 [1 possibility]

Total from q=0 case: 1 + 3 + ... + 15 + 17 = 81

Case q=1 [ n + 2d ≤ 11 ]

Case (q=1, d=0) → n = 0 through 11 [12]

Case (q=1, d=1) → n = 0 through 9 [10]

...

Case (q=1, d=5) → n = 0 through 1 [2]

Total from q=1 case: 2 + 4 + ... + 10 + 12 = 42

Case q=2 [ n + 2 ≤ 6 ]

Case (q=2, d=0) → n = 0 through 6 [7]

Case (q=2, d=1) → n = 0 through 4 [5]

Case (q=2, d=2) → n = 0 through 2 [3]

Case (q=2, d=3) → n = 0 [1]

Total from case q=2: 1 + 3 + 5 + 7 = 16

Case q=3 [ n + 2d ≤ 1 ]

Here d must be 0, so there is only the case:

Case (q=3, d=0) → n = 0 through 1 [2]

So the case q=3 only has 2.

Grand total: 2 + 16 + 42 + 81 = 141