# 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 B

All the unknown entries can be expressed in terms of $b$. Since $100e = beh = ceg = def$, it follows that $h = \frac{100}{b}, g = \frac{100}{c}$, and $f = \frac{100}{d}$. Comparing rows $1$ and $3$ then gives $50bc = 2 \cdot \frac{100}{b} \cdot \frac{100}{c}$, from which $c = \frac{20}{b}$. Comparing columns $1$ and $3$ gives $50d \cdot \frac{100}{c}= 2c \cdot \frac{100}{d}$, from which $d = \frac{c}{5} = \frac{4}{b}$. Finally, $f = 25b, g = 5b$, and $e = 10$. All the entries are positive integers if and only if $b = 1, 2,$ or $4$. The corresponding values for $g$ are $5, 10,$ and $20$, and their sum is $\boxed{\mathbf{(C)}35}$.

Credit to Solution B goes to http://billingswest.billings.k12.mt.us/math/AMC%201012/AMC%2012%20work%20sheets/2004%20AMC%2012B%20ws-15.pdf, a page with a play-by-play explanation of the solutions to this test's problems.

## Solution C

We know because this is a multiplicative magic square that each of the following are equal to each other: $100e=ceg=50dg=beh=2cf=50bc=def=2gh$

From this we know that $50dg=2hg$, thus $h=25d$. Thus $beh=be(25d)$ and $be(25d)=100e$. Thus $b=\frac{4}{d}$ From this we know that $50bc=(50)(\frac{4}{d})(c)=50dg$. Thus $c=\frac{d^2g}{4}$. Now we know from the very beginning that $100e=ceg$ or $100=cg$ or $100=\frac{d^2g}{4}(g)$ or $\frac{d^2g^2}{4}$. Rearranging the equation $100=\frac{d^2g^2}{4}$ we have $(d^2)(g^2)=400$ or $dg=20$ due to $d$ and $g$ both being positive. Now that $dg=20$ we find all pairs of positive integers that multiply to $20$. There is $(d,g)= (20,1);(10,2);(5,4);(4,5);(2,10);(1,20)$. Now we know that $b=\frac{4}{d}$ and b has to be a positive integer. Thus $d$ can only be $1$, $2$, or $4$. Thus $g$ can only be $20$, $10$,or $5$. Thus sum of $20+10+5$ = $35$. The answer is $\boxed{\mathbf{(C)}35}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 