Solution to Cookie Seller

The solution goes lik this.. :)
We hav inifinte number of boxes of each type..
By taking all multiples of 6 we will get all the even multiples of 3 above 6.. and by taking all multiples of 9 we will be able to make all odd multiples of 3. Thus we can make all the multiples of 3 above the number 6..
So wat abt the the other numbers?? hmmm..
Now take the number 29.. it gives 2 as remainder when divided by 3.. if we add 400 to 29 ie 429 is a multiple of 3. So any number above 410 which gives a remainder of 2 when divided by 3 can be made by using a 400 box and the rest 6 and 9.
ie, 421= can be made as 400 + 21 ( 2 *6+1*9) and so on..
Now take the number 28. it gives a remainder 1 when divided by 3. If we add 800 to 28 ie 828 is divisble by 3..
So any number above 806 can be made by these 3 types of boxes..
So wat abt the maximum tat cannot be sold
805 - 400 is divisible by 3..
804 is divisible by 3
803 is nt divisble by 3, 803-400=403 is nt divisble by 3, a 803-800=3 cannot be made using our boxes..

So 803 is the max number of boxes tat cannot be sold..