Solution for #20You can buy any number of nuggets 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. 
