Suppose there is a pile of quarters, dimes, and pennies with a total value of $1.06. How much of each coin can be present without being able to make change for a dollar? If there are multiple selections of coins that will work, choose the selection with the largest total number of coins.

Hello,

I note (a, b, c) the result of a quarters, b dimes and c pennies:

2 solutions:

106 = (3, 3, 1) = (1, 8, 1)

106 = (0, 0, 106) but : 100 = 0*25 + 0*10 + 100

106 = (0, 1, 96) but : 100 = 0*25 + 1*10 + 90

106 = (0, 2, 86) but : 100 = 0*25 + 2*10 + 80

106 = (0, 3, 76) but : 100 = 0*25 + 3*10 + 70

106 = (0, 4, 66) but : 100 = 0*25 + 4*10 + 60

106 = (0, 5, 56) but : 100 = 0*25 + 5*10 + 50

106 = (0, 6, 46) but : 100 = 0*25 + 6*10 + 40

106 = (0, 7, 36) but : 100 = 0*25 + 7*10 + 30

106 = (0, 8, 26) but : 100 = 0*25 + 8*10 + 20

106 = (0, 9, 16) but : 100 = 0*25 + 9*10 + 10

106 = (0, 10, 6) but : 100 = 0*25 + 10*10 + 0

106 = (1, 0, 81) but : 100 = 1*25 + 0*10 + 75

106 = (1, 1, 71) but : 100 = 1*25 + 1*10 + 65

106 = (1, 2, 61) but : 100 = 1*25 + 2*10 + 55

106 = (1, 3, 51) but : 100 = 1*25 + 3*10 + 45

106 = (1, 4, 41) but : 100 = 1*25 + 4*10 + 35

106 = (1, 5, 31) but : 100 = 1*25 + 5*10 + 25

106 = (1, 6, 21) but : 100 = 1*25 + 6*10 + 15

106 = (1, 7, 11) but : 100 = 1*25 + 7*10 + 5

106 = (1, 8, 1) is good

106 = (2, 0, 56) but : 100 = 2*25 + 0*10 + 50

106 = (2, 1, 46) but : 100 = 2*25 + 1*10 + 40

106 = (2, 2, 36) but : 100 = 2*25 + 2*10 + 30

106 = (2, 3, 26) but : 100 = 2*25 + 3*10 + 20

106 = (2, 4, 16) but : 100 = 2*25 + 4*10 + 10

106 = (2, 5, 6) but : 100 = 2*25 + 5*10 + 0

106 = (3, 0, 31) but : 100 = 3*25 + 0*10 + 25

106 = (3, 1, 21) but : 100 = 3*25 + 1*10 + 15

106 = (3, 2, 11) but : 100 = 3*25 + 2*10 + 5

106 = (3, 3, 1) is good

106 = (4, 0, 6) but : 100 = 4*25 + 0*10 + 0