Listing is never away of counting as you might of missed some. Now having said that, listing is good to sometimes see how to count something.
Generally this is true, but listing is a fine tactic for this problem as there's only finitely many solutions, so we can definitely generate an exhaustive list. Given the condition that
5r+4s+3t+6u=100 and knowing that all four variables must be non-negative integers, we can immediately deduce that
r≤20. Similarly, we can deduce that
s≤25. However, the additional condition that
r≥s≥t≥u ≥0 means we can discard any solutions where
s≥21... and so on.
I'll stop there for brevity's sake, but it should be clear how this process of logical deductions can be continued to generate a full list of every possibility. I'm not quite sure where Denis went wrong with his second attempt, but there are actually a total of
29 solutions:
- 5(7) + 4(5) + 3(5) + 6(5) = 100
- 5(8) + 4(6) + 3(4) + 6(4) = 100
- 5(8) + 4(6) + 3(6) + 6(3) = 100
- 5(9) + 4(7) + 3(3) + 6(3) = 100
- 5(9) + 4(7) + 3(5) + 6(2) = 100
- 5(9) + 4(7) + 3(7) + 6(1) = 100
- 5(10) + 4(5) + 3(4) + 6(3) = 100
- 5(10) + 4(8) + 3(2) + 6(2) = 100
- 5(10) + 4(8) + 3(4) + 6(1) = 100
- 5(10) + 4(8) + 3(6) + 6(0) = 100
- 5(11) + 4(6) + 3(3) + 6(2) = 100
- 5(11) + 4(6) + 3(5) + 6(1) = 100
- 5(11) + 4(9) + 3(1) + 6(1) = 100
- 5(11) + 4(9) + 3(3) + 6(0) = 100
- 5(12) + 4(4) + 3(4) + 6(2) = 100
- 5(12) + 4(7) + 3(2) + 6(1) = 100
- 5(12) + 4(7) + 3(4) + 6(0) = 100
- 5(12) + 4(10) + 3(0) + 6(0) = 100
- 5(13) + 4(5) + 3(3) + 6(1) = 100
- 5(13) + 4(5) + 3(5) + 6(0) = 100
- 5(13) + 4(8) + 3(1) + 6(0) = 100
- 5(14) + 4(3) + 3(2) + 6(2) = 100
- 5(14) + 4(6) + 3(2) + 6(0) = 100
- 5(15) + 4(4) + 3(1) + 6(1) = 100
- 5(15) + 4(4) + 3(3) + 6(0) = 100
- 5(16) + 4(2) + 3(2) + 6(1) = 100
- 5(16) + 4(5) + 3(0) + 6(0) = 100
- 5(17) + 4(3) + 3(1) + 6(0) = 100
- 5(20) + 4(0) + 3(0) + 6(0) = 100
Thus, the maximum sum is 10 + 8 + 6 + 0 = 24 (or 9 + 7 + 7 + 1 = 24), and the minimum sum is 20 + 0 + 0 + 0 = 20