He must have an odd number of 5's, otherwise
the sum would be a multiple of 10. But he
has to have more than just 1 five, because
there is no way to make the remaining 120
with 12 bills other than with 12 10's, but
that would be using no 20's. So
1. If he has 3 5's, he must make the remaining
110 with 10 tens and twenties.
3 fives, 9 tens, and 1 twenty.
3+9+1 = 13
(3)(5)+(9)(10)+1(20) = 15+90+20 = 125
or
2. If he has 5 5's, he must make the remaining
100 with 8 tens and twenties.
5 fives, 6 tens, and 2 twenties.
5+6+2 = 13
(5)(5)+(6)(10)+2(20) = 25+60+40 = 125
or
3. If he has 7 5's, he must make the remaining
90 with 6 tens and twenties.
7 fives, 3 tens, and 3 twenties.
7+3+3 = 13
(7)(5)+(3)(10)+3(20) = 35+30+60 = 125
Edwin
