10d + 25q = 130
Divide through by 5
2d + 5q = 26
2 is the least coefficient in absolute value,
and 26 is already a multiple of 2, so we write
5 in terms of its nearest multiple of 2
2d + (4+1)q = 26
2d + 4d + q = 26
d + 2q + q/2 = 13
Isolate the fraction:
q/2 = 13 - d - 2q
The right side is an integer and the left side
is non-negative, so both sides equal some non-
negative integer A.
q/2 = A and 13 - d - 2q = A
q = 2A, substituting in
13 - d - 2q = A
13 - d - 2(2A) = A
13 - d - 4A = A
13 - 5A = d
The minimum number of dimes is 0
and the maximum number of dimes is 13,
so 0 ≦ d ≦ 13
0 ≦ 13 - 5A ≦ 13
-13 ≦ -5A ≦ 0
2.6 ≧ A ≧ 0
The minimum number of quarters is 0
and the maximum number of quarters is 5,
so 0 ≦ q ≦ 5
0 ≦ 2A ≦ 5
0 ≦ A ≦ 2.5
So A = 0, A = 1 or A = 2
A 13-5A = d 2A = q
0 13-5(0) = 13 2(0) = 0
1 13-5(1) = 8 2(1) = 2
2 13-5(2) = 3 2(2) = 4
So
13 dimes and no quarters.
8 dimes and 2 quarters.
3 dimes and 4 quarters.
Edwin