# Problem of the Week Problem C and Solution Odd Boxes

## Problem

The first $$9$$ positive odd integers are placed in the following $$3$$ by $$3$$ grid in such a way that the sum of the numbers in each row, column and main diagonal is the same. Four of the numbers are shown and the other five numbers are represented by the letters $$A$$, $$B$$, $$C$$, $$D$$, and $$E$$.

Determine the values of $$A$$, $$B$$, $$C$$, $$D$$, and $$E$$.

$\begin{array}{|c|c|c|} \hline A & 5 & B \\\hline C & D & 17 \\\hline 11 & 13 & E \\\hline \end{array}$

## Solution

The final answer is $$A= 15$$, $$B=7$$, $$C=1$$, $$D=9$$, and $$E=3$$, which we will justify below in two different ways.

Solution 1

The numbers to be placed in the grid are $$1$$, $$3$$, $$5$$, $$7$$, $$9$$, $$11$$, $$13$$, $$15$$, and $$17$$, the first $$9$$ positive odd integers. Therefore, the sum of all of the numbers in the grid is $$1+3+5+7+9+11+13+15+17=81$$. It follows that the sum of the sums of the three rows is $$81$$. But each row has the same sum, so the sum of the numbers in each row is $$81\div 3=27$$. We know that the sum of the numbers in each row, column and diagonal is the same. Therefore, the sum of the numbers in each column is also equal to $$27$$ and the sum of the numbers in each diagonal is also equal to $$27$$.

We can now use this information to determine the values in each cell of the $$3$$ by $$3$$ grid. In the third row, we know that $$11+13+E=27$$ or $$24+E=27$$ and $$E=3$$ follows. Here is the updated grid: $\begin{array}{|c|c|c|} \hline A & 5 & B \\\hline C & D & 17 \\\hline 11 & 13 & 3 \\\hline \end{array}$

In the second column, we know that $$5 + D + 13 = 27$$ and $$D=9$$ follows. Here is the updated grid: $\begin{array}{|c|c|c|} \hline A & 5 & B \\\hline C & 9 & 17 \\\hline 11 & 13 & 3 \\\hline \end{array}$

In second row, we know that $$C+9+17=27$$ and $$C= 1$$ follows. Here is the updated grid: $\begin{array}{|c|c|c|} \hline A & 5 & B \\\hline 1 & 9 & 17 \\\hline 11 & 13 & 3 \\\hline \end{array}$

In the first column, we know that $$A + 1 + 11 =27$$ and $$A = 15$$ follows. Here is the updated grid: $\begin{array}{|c|c|c|} \hline 15 & 5 & B \\\hline 1 & 9 & 17 \\\hline 11 & 13 & 3 \\\hline \end{array}$

In the first row, we know that $$15 + 5 + B=27$$ and $$B=7$$ follows. Here is the updated grid: $\begin{array}{|c|c|c|} \hline 15 & 5 & 7 \\\hline 1 & 9 & 17 \\\hline 11 & 13 & 3 \\\hline \end{array}$

Therefore, $$A= 15$$, $$B=7$$, $$C=1$$, $$D=9$$, and $$E=3$$. With these values, we can see that each of the first $$9$$ positive odd integers appears in the grid, and indeed, the sum of the numbers in each row, column, and main diagonal is the same.

Solution 2

In this solution we will determine the unknown values without finding that the row, column, and diagonal sum is $$27$$. This solution will use more algebra.

Since the sum of the numbers in the third row is equal to the sum of the numbers in the third column, we know that \begin{aligned} 11 + 13 + E &= B + 17 + E\\ 11 + 13 &=B + 17\\ B&=7\end{aligned}

Again, since the sum of the numbers in the first row is equal to the sum of the numbers in the first column, we know that \begin{aligned} A+5+7&=A+C+11\\ 5+7&=C+11\\ C&=1\end{aligned}

Here is the updated grid: $\begin{array}{|c|c|c|} \hline A & 5 & 7 \\\hline 1 & D & 17 \\\hline 11 & 13 & E \\\hline \end{array}$

We also know that the two diagonals have the same sum, so we have \begin{aligned} A+D+E&=7+D+11\\ A+E &=18 \end{aligned}

We have used the odd numbers $$1$$, $$5$$, $$7$$, $$11$$, $$13$$, and $$17$$. This leaves the odd numbers $$3$$, $$9$$ and $$15$$ for $$A$$, $$D$$, and $$E$$. Since $$A + E = 18$$, $$A$$ and $$E$$ must be $$3$$ and $$15$$, in some order.

If $$A = 3$$ and $$E=15$$, then the sum of the first row is $$3 + 5 + 7=15$$ and the sum of the third row is $$11 + 13 + 15 = 39$$. These sums are not the same. Therefore, $$A=3$$ and $$E=15$$ is not correct.

If $$A=15$$ and $$E=3$$, then the sum of the first row is $$15 + 5 + 7=27$$ and the sum of the third row is $$11 + 13 + 3 = 27$$. These sums are the same. Therefore, $$A=15$$ and $$E=3$$. This leaves $$D=9$$.

Therefore, $$A= 15$$, $$B=7$$, $$C=1$$, $$D=9$$, and $$E=3$$. From here, one can easily show each row, column, and diagonal sums to 27, as we found in Solution 1.