Kane's bakery makes four flavours of pie. The weight of each flavour pie is follows: apple pie - 700 grams, banoffee pie - 725 grams, custard pie - 600 grams, rhubarb pie - 675 grams. Robert makes an order for 4 apple pies, 8 banoffee pies, 3 custard pies and 5 rhubarb pies to sell at a fundraiser. Kane packs the order into boxes that each hold 4 kilograms. What is the fewest number of boxes he can pack the order into?

It is a very difficult problem if the box sizes are smaller or if there are more pies to pack.

For this particular problem, we first calculate the total weight of pies:

4 apple pies at 700 g = 2800 g

8 banoffee pies at 725 g = 5800 g

3 custaard pies at 600 g = 1800 g

5 rhubarb pies at 675 g = 3375 g

for a total of 13775 g

The minimum number of boxes needed will be 13775/4000=3.44, or

at least 4 boxes will be required.

We also note that 4 boxes will hold 16000 g, and we only need to pack 13775g, so there is a lot of margin.

We can proceed to do the mechanical way, fill in order of apple, banoffee, custard, and finally rhubarb pies.

Following table shows how we fill, naively:

box content Remaining to pack

Box A B C R Weight A B C R

0 4 8 3 5 (original order)

1 4 1 0 0 3525 0 7 3 5

2 0 5 0 0 3625 0 2 3 5

3 0 2 3 0 3250 0 0 0 5

4 0 0 0 5 3375 0 0 0 0

Total weight 13775 g

So that does the job, 4 boxes (minimum) with none exceeding 4000 g to pack a total of 13775 g of pies.