2. PB and J Array Ken has made a selection of square jam sandwiches for party, which he arranges in a single layer on the table
in perfect rectangle. Sarah has brought along the same number of square peanut butter sandwiches, which she puts out, also as a single layer, to form a border vA (one sandwich wide) around Ken's Given that all of the sandwiches are the same size and they are all used, how many sandwiches could each person possibly have made?
2 answers:
Let's assume Ken arranged his sandwiches in a rectangular shape with length 'L' and width 'W'. Now Sarah arranged her sandwiches to form a border around Ken's rectangle. The solution diagram is attached below.
Ken needs as much sandwiches as area of rectangle i.e. number of Ken's sandwiches would be = LW.
It is clear that Sarah needs as much sandwiches as perimeter of rectangle and extra four sandwiches to put on corners. It means number of Sarah's sandwiches would be = 2·(L+W) + 4.
Given that all of the sandwiches are the same size and they are all used.
Ken and Sarah bought same number of sandwiches. So we need to solve the following equation :-
Number of Ken's sandwiches = Number of Sarah's sandwiches
LW = 2·(L+W) + 4
Using trial and error approach, we get 6 x 4 = 2 x (6 + 4) + 4.
So number of Ken's sandwiches would be LW = 6 x 4 = 24 sandwiches.
Hence, final answer is "Each person could have 24 sandwiches".
You might be interested in
The rows add up to
The sum of the numbers in row
The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When
so the base case holds. Assume the claim holds for
Use this to show that it holds for
Notice that
So you can write the expansion for
and since
and so the claim holds for
Setting
which agrees with the result obtained for part (c).