2004 AMC 12B Problems/Problem 22

Problem

The square

$\begin{tabular}{|c|c|c|} \hline 50 & \textit{b} & \textit{c} \\ \hline \textit{d} & \textit{e} & \textit{f} \\ \hline \textit{g} & \textit{h} & 2 \\ \hline \end{tabular}$

is a multiplicative magic square. That is, the product of the numbers in each row, column, and diagonal is the same. If all the entries are positive integers, what is the sum of the possible values of $g$ ?

$\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 62 \qquad \textbf{(E)}\ 136$

Solution

If the power of a prime $p^n$ other than $2,5$ divides $g$ , then from $50 \cdot 2 e = 50dg$ it follows that $p^n|e$ , but then considering the product of the diagonals, $p^{2n} |gec$ but $p^{2n} \nmid 100e$ , contradiction. So the only prime factors of $g$ are $2$ and $5$ .

It suffices now to consider the two magic squares comprised of the powers of $2$ and $5$ of the corresponding terms. These satisfy the normal requirement that the sums of rows, columns, and diagonals are the same, owing to our rules of exponents; additionally, all terms are non-negative.

The powers of $2$ :

$\begin{tabular}{|c|c|c|} \hline 1 & \textit{b} & \textit{c} \\ \hline \textit{d} & \textit{e} & \textit{f} \\ \hline \textit{g} & \textit{h} & 1 \\ \hline \end{tabular}$

So $1 + 1 + e = g + e + c \Longrightarrow g = 2 - c$ , so $g = 0,1,2$ . Indeed, we have the magic squares

$\begin{tabular}{|c|c|c|} \hline 1 & 0 & 2 \\ \hline 2 & 1 & 0 \\ \hline 0 & 2 & 1 \\ \hline \end{tabular}, \quad \begin{tabular}{|c|c|c|} \hline 1 & 1 & 1 \\ \hline 1 & 1 & 1 \\ \hline 1 & 1 & 1 \\ \hline \end{tabular}, \quad \begin{tabular}{|c|c|c|} \hline 1 & 2 & 0 \\ \hline 0 & 1 & 2 \\ \hline 2 & 0 & 1 \\ \hline \end{tabular},$

The powers of $5$ :

$\begin{tabular}{|c|c|c|} \hline 2 & \textit{b} & \textit{c} \\ \hline \textit{d} & \textit{e} & \textit{f} \\ \hline \textit{g} & \textit{h} & 0 \\ \hline \end{tabular}$

Again, we get $2 + e = g + e + c \Longrightarrow g = 0,1,2$ . However, if we let $g = 2, c = 0$ , then $e = d + e + f \Longrightarrow d = f = 0$ , which obviously gives us a contradiction, and similarly for $g = 0, c = 2$ . For $g = 1$ , we get

$\begin{tabular}{|c|c|c|} \hline 2 & 0 & 1 \\ \hline 0 & 1 & 2 \\ \hline 1 & 2 & 0 \\ \hline \end{tabular}$

In conclusion, $g$ can be $2^0 \cdot 5^1, 2^1 \cdot 5^1, 2^2 \cdot 5^1$ , and their sum is $\boxed{\mathbf{(C)}35}$ .

2004 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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2004 AMC 12B Problems/Problem 22

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