### The Frog Problem

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.

- How many packages are there?
- 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
so

B + C = B + (B + 1) = 2B + 1
an odd number. From this we conclude that

B + C = 9
or

2B + 1 = 9
or

2B = 8
or

B = 4
So

C = B + 1 = 5
and if

A + B = 7
then

A + 4 = 7
and

A = 3
If

C + D = 12
then

5 + D = 12
and

D = 7
So the weights are

3, 4, 5, 7