5. Put 1 cent, 2 cent, 5 cent, and 10 cent coins into 19 boxes, so that each box contains coins, and the amount of money in any two boxes is different. How many coins are needed at least? At this time, what is the minimum total amount of money in all the boxes?
5. Solution: Taking one coin, there are 4 types: 1 cent, 2 cents, 5 cents, 10 cents (1 jiao);

Taking two coins, we have $1+1=2$ (cents), $2+2=4$ (cents), $5+5=10$ (cents), $10+10=20$ (cents) (2 jiao),
$1+2=3$ (cents), $1+5=6$ (cents), $1+10=11$ (cents) (1 jiao 1 cent), $2+5=7$ (cents), $2+10=12$ (cents) (1 jiao 2 cents), $5+10=15$ (cents) (1 jiao 5 cents),

There are 10 types in total, with 2 types repeated (2 cents, 10 cents). Adding the 4 types from taking one coin, there are 12 different coin values;

Taking three coins, we can add 1 cent, 2 cents, 5 cents, and 10 cents to the 10 types from taking two coins, resulting in 40 types. From smallest to largest, 7 non-repeating coin values are: 8 cents, 9 cents, 13 cents, 14 cents, 16 cents, 17 cents, 21 cents, adding to the 12 types above, there are 19 types in total. The total number of coins used is: $1 \times 4 + 2 \times 8 + 3 \times 7 = 41$ (coins). The total amount of money is: $1+2+3+\cdots+17+20+21=194$ (cents)