Math 300


The Frog Problem

Dr. Wilson

June is a biology teacher. She gets a shipment of dead frogs for her students to disect, but when she opens them up, she finds that they have gone bad and can't be used. She calls up the biology supply warehouse and complains. They tell her that they need to know how much the packages of dead frogs weigh. They tell her that they always package the dead frogs in packages that weigh a whole number of pounds. When she goes to weigh the frogs, she finds that the smallest weight that her scale can weigh is 7 pounds, and most of the packages weigh less than that, so she weighs the packages in groups of two. She forms every possible group of two packages and gets the weights 7, 8, 9, 10, 11, and 12 pounds.

  1. How many packages are there?
  2. What are the weights of the packages?

First we have to determine how many packages there are. For this we can use Pascal's triangle. When two packages are weighed, the two packages which are being weighed form a 2 element subset of the 4 element set of packages. When we look at the numbers of two element subsets in Pascal's triangle,


we see that there are 6 two element subsets of a 4 element set, so there must be 4 packages. Let us let A, B, C, and D denote the weights of the packages. All of the packages must have different weights because if A and B were the same weights, then A + C would weight the same amount as B + C, which doesn't happen because all of the weights of the pairs of packages are different. Let us suppose that the packages are labled so that A is the lightest, B is the next lightest, and D is the heaviest. Then A + B would be the lightest pair of packages, and A + C would be the next lightest. D + C would be the heaviest, and D + B would be the next heaviest. This gives us the equations

A + B = 7

A + C = 8

D + B = 11

D + C = 12

There are two other subsets, A + D, and B + C. One of these will weith 9 and the other will weigh 10, but it will take a little more reasoning to determine which is which.

If we subtract the first equation from the second we get

C - B = 1

We get the same thing if we subtract the third equation from the 4th. This tells us that

C = B + 1


B + C = B + (B + 1) = 2B + 1

an odd number. From this we conclude that

B + C = 9


2B + 1 = 9


2B = 8


B = 4


C = B + 1 = 5

and if

A + B = 7


A + 4 = 7


A = 3


C + D = 12


5 + D = 12


D = 7

So the weights are

3, 4, 5, 7