Suppose there is a pile of quarters dimes and pennies with a total value of $1.07 how much of each coin can be present without being able to make change for a dollar? Of there are multiple selections of the coins that will work choose the selection with the largest total numbers of coins.

Hello,

Very nice as problem.

2 solutions:

1 quater, 8 dimes, 2 pennies

and

3 quaters, 3 dimes, 2 pennies

since

107 = (0, 0, 107) but : 100 = 0*25 + 0*10 + 100

107 = (0, 1, 97) but : 100 = 0*25 + 1*10 + 90

107 = (0, 2, 87) but : 100 = 0*25 + 2*10 + 80

107 = (0, 3, 77) but : 100 = 0*25 + 3*10 + 70

107 = (0, 4, 67) but : 100 = 0*25 + 4*10 + 60

107 = (0, 5, 57) but : 100 = 0*25 + 5*10 + 50

107 = (0, 6, 47) but : 100 = 0*25 + 6*10 + 40

107 = (0, 7, 37) but : 100 = 0*25 + 7*10 + 30

107 = (0, 8, 27) but : 100 = 0*25 + 8*10 + 20

107 = (0, 9, 17) but : 100 = 0*25 + 9*10 + 10

107 = (0, 10, 7) but : 100 = 0*25 + 10*10 + 0

107 = (1, 0, 82) but : 100 = 1*25 + 0*10 + 75

107 = (1, 1, 72) but : 100 = 1*25 + 1*10 + 65

107 = (1, 2, 62) but : 100 = 1*25 + 2*10 + 55

107 = (1, 3, 52) but : 100 = 1*25 + 3*10 + 45

107 = (1, 4, 42) but : 100 = 1*25 + 4*10 + 35

107 = (1, 5, 32) but : 100 = 1*25 + 5*10 + 25

107 = (1, 6, 22) but : 100 = 1*25 + 6*10 + 15

107 = (1, 7, 12) but : 100 = 1*25 + 7*10 + 5

107 = (1, 8, 2) is good

107 = (2, 0, 57) but : 100 = 2*25 + 0*10 + 50

107 = (2, 1, 47) but : 100 = 2*25 + 1*10 + 40

107 = (2, 2, 37) but : 100 = 2*25 + 2*10 + 30

107 = (2, 3, 27) but : 100 = 2*25 + 3*10 + 20

107 = (2, 4, 17) but : 100 = 2*25 + 4*10 + 10

107 = (2, 5, 7) but : 100 = 2*25 + 5*10 + 0

107 = (3, 0, 32) but : 100 = 3*25 + 0*10 + 25

107 = (3, 1, 22) but : 100 = 3*25 + 1*10 + 15

107 = (3, 2, 12) but : 100 = 3*25 + 2*10 + 5

107 = (3, 3, 2) is good

107 = (4, 0, 7) but : 100 = 4*25 + 0*10 + 0