Impossible Combination

At John's Bakery, you can order Milky breads in boxes of 6, 9, and 20. What is the largest number of breads that it is not possible to obtain by purchasing some combination of these boxes?

Answer:

You can buy any number of breads that is evenly divisible by three except for three just by using combinations of boxes of 6 and boxes of 9. (Use at most one box of 9, then multiples of boxes of 6.). If the number is not divisible by three, use a box of 20. If, after a box of 20 is purchased, the remaining number is divisible by three, you're all set. Otherwise, use a second box of 20. The remaining number will necessarily be divisible by 3, and you're all set. So the largest number that cannot be purchased would be one that requires two boxes of 20 before the remainder is reduced to a number divisible by three. Since three is the only number evenly divisible by three that cannot be purchased, the largest impossible number is 3+ 20 + 20 = 43.