2004 AMC 12B Problems/Problem 22

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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 A

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 $\boxed{\mathbf{(C)}35}$.

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.

See also

2004 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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