Math Problem Need help!

How many ways can eight dollars be changed into dimes and/or quarters?


please explain :)

Answer: 17.
Let A denotes number of dimes and B number of quarters (A and B are non-negative integers). This gives us equation
10*A + 25*B = 800 (:5)
2*A + 5*B = 160
Thus B = 32 - (2*A):5. As B is an integer, (2A)/5 must be integer. That gives 2*A is a multiply of 5, thus A is a multiply of 5 so can be expressed in the form A = 5*C where C is integer, C >= 0.
Then B = 32 - 2*C and B >= 0 gives us C <= 16.

So every way to change has the form (5*C dimes, 32-2*C quarters) and total number of ways is equaled to number of choosing integer 0 <= C <= 16, totally 17.

According to my knowledge. One can use casework.
For example.
0 dimes, 32 quarters.
1 dime not possible.
..... 2, 3, 4 dimes not possible.
5 dimes, 30 quarters.
10 dimes, 28 quarters.
15 dimes
...........
75 dimes, 2 quarters.
80 dimes, 0 quarters.
i believe there are 17 ways

Just go down the list

1) All Dimes
2) 7.50 dimes, .50 quarters
3) 7.00 dimes, 1.00 quarters
4) 6.50 dimes, 1.50 quarters
.
.
.
n) All Qurters