# Can Blaine distribute 15 money bags fairly?

## Backstory

Blaine Fairway, a senior manager at CurrFlow Industries, Inc., has been personally accepting large rebates for multi-billion-dollar purchases from certain manufacturers for years, although that money legally should have gone to his company.

## Extortion

Recently, 15 of his subordinates found out about Blaine's illegal activities, but instead of reporting his crimes to the company, they decided as a group that they are entitled to most of that money. So, they demanded that the past 15 rebate payments be distributed to them fairly among all their teams, or else they would report him. Fairly means that the average of the amounts of money, for the subordinates on a particular team, is the same for each team. There are 9 teams, as detailed below. Each subordinate is on 2 teams.

## Problem

Blaine still has each of the past 15 payments, which he has stored in small money bags. He is willing to pay them off, but he cannot figure out a fair way to distribute the money to them. He knows that the average amount of money on each of the 9 teams must be the same, or they will be angry and surely report him.

## Question

Is it possible for Blaine to distribute the 15 money bags (below), 1 to each subordinate, such that it is fair? If so, give a possible distribution. If not, prove it.

Team A: Abe, Ally, Abbie, Aaron
Team B: Bo, Brie, Ben, Bob, Brianna
Team C: Chris, Cedric, Candy, Cindy, Cam, Candace
Team 1: Abe, Bo, Chris
Team 2: Ally, Brie, Cedric
Team 3: Abbie, Ben, Candy
Team 4: Aaron, Bob, Cindy
Team 5: Brianna, Cam
Team 6: Candace


The newly formulated question stipulates that the average amount of money on each team must be the same — i.e. \$80,000 per employee. This is a much less trivial problem than the previous formulation, as now there is a plausible starting point. We can formulate the problem now as a grid sum puzzle: \begin{matrix} A_1 & B_1 & C_1 & 240 \\ A_2 & B_2 & C_2 & 240 \\ A_3 & B_3 & C_3 & 240 \\ A_4 & B_4 & C_4 & 240 \\ & B_5 & C_5 & 160 \\ & & C_6 & 80 \\ 320 & 400 & 480 & \end{matrix} Where the numbers in each row add up to the number on the right, the numbers on each column add up to the number below, and the numbers to be filled in are the values of the bags. Firstly, Candace on her own team must receive the$80,000 bag.

\begin{matrix} A_1 & B_1 & C_1 & 240 \\ A_2 & B_2 & C_2 & 240 \\ A_3 & B_3 & C_3 & 240 \\ A_4 & B_4 & C_4 & 240 \\ & B_5 & C_5 & 160 \\ & & 80 & 80 \\ 320 & 400 & 480 & \end{matrix}

Next, Brianna and Cam must receive a total of \$160,000, so their only options are a 69/91 or a 63/97 split. We can't fill in the table with this information yet, but we can find four sets of three numbers that add up to 240. Remaining: 55 63 65 68 69 81 82 85 87 88 89 91 97 100  We've deduced that the \$100K bag must be in a triplet, with the other two bags summing to \$140,000. The only pair that satisfies this is 55 and 85. Remaining: 63 65 68 69 81 82 87 88 89 91 97  Next, we try the \$97K bag, with the other two bags summing to \$143,000. There's no pair in the remaining bags that satisfies this, so we deduce that Brianna and Cam must have a 63/97 split, and move on to the next set of bags. Remaining: 65 68 69 81 82 87 88 89 91  Next, the \$91K bag, with the other two bags summing to \$149,000. We find that 68 and 81 satisfy this one. Remaining: 65 69 82 87 88 89  Next, the \$89K bag, with the other two bags summing to \$151,000. We find that 69 and 82 satisfy this one. Remaining: 65 87 88  And finally, the last three remaining bags sum up to \$240,000 as required.

So now we've got all the rows, but we need to arrange them so that the columns sum up as well.

\begin{matrix} 55 & 85 & 100 & 240 \\ 65 & 87 & 88 & 240 \\ 68 & 81 & 91 & 240 \\ 69 & 82 & 89 & 240 \\ & 63 & 97 & 160 \\ & & 80 & 80 \\ 320? & 400? & 480? & \end{matrix}

First, we want to try to get the leftmost column (team A) to sum up to 320. The sum $85 + 65 + 81 + 89$ works.

\begin{matrix} 85 & 55 & 100 & 240 \\ 65 & 87 & 88 & 240 \\ 81 & 68 & 91 & 240 \\ 89 & 82 & 69 & 240 \\ & 63 & 97 & 160 \\ & & 80 & 80 \\ 320 & 400? & 480? & \end{matrix}

Then, to get the middle column (team B) to sum up to 400 using only numbers from the middle column to the right column. The sum $100 + 87 + 68 + 82 + 63$ works.

\begin{matrix} 85 & 100 & 55 & 240 \\ 65 & 87 & 88 & 240 \\ 81 & 68 & 91 & 240 \\ 89 & 82 & 69 & 240 \\ & 63 & 97 & 160 \\ & & 80 & 80 \\ 320 & 400 & 480 & \end{matrix}

And the remaining numbers will naturally sum to 480, as expected. There are more possible configurations, involving swapping the top four rows or the bottom two numbers, but this is one of the possibilities.

So, the money is distributed as follows:

Aaron receives \$89,000. Abbie receives \$81,000.
Abe receives \$85,000. Ally receives \$65,000.
Ben receives \$68,000. Bo receives \$100,000.
Bob receives \$82,000. Brianna receives \$63,000.
Brie receives \$87,000. Cam receives \$97,000.
Candace receives \$80,000. Candy receives \$91,000.
Cedric receives \$88,000. Chris receives \$55,000.