Ken has made a selection of square jam sandwiches for a party, which he arranges in a single layer on the table in a 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 (one sandwich wide) around Ken's rectangle. Given that all of the sandwiches are the same size and they are all used, how many sandwiches could each person possibly have?

9514 1404 393Answer:   24 or 30Step-by-step explanation:If the rectangle dimensions are x by y, then there are xy sandwiches in the rectangle and (x+2)(y+2)-xy sandwiches in the border. If those are the same numbers, we require ...   xy = (x+2)(y+2)-xy   2xy = xy +2x +2y +4   xy = 2(x +y +2)There are no integer solutions for x=1 or 2. For x=3, we have ...   3y = 2(3 +y +2)   3y = 2y +10   y = 10 . . . . . . . . 3·10 = 30 sandwiches will make a 3×10 rectangle__For x=4, we have ...   4y = 2(4 +y +2)   4y = 2y +12   y = 6 . . . . . . . 4·6 = 24 sandwiches will make a 4×6 rectangle__There are no other integer solutions.Ken and Sarah could each have 24 or 30 sandwiches.