Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2004 AMC 12B Problems/Problem 22"

(→‎Solution)
m (→‎Solution B)
Line 53: Line 53:
  
 
==Solution B==
 
==Solution B==
−
All the unknown entries can be expressed in terms of b.
+
All the unknown entries can be expressed in terms of <math>b</math>.
−
Since 100e = beh = ceg = def, it follows that h = 100/b, g = 100/c,
+
Since <math>100e = beh = ceg = def</math>, it follows that <math>h = \frac{100}{b}, g = \frac{100}{c}</math>,
−
and f = 100/d. Comparing rows 1 and 3 then gives
+
and <math>f = \frac{100}{d}</math>. Comparing rows <math>1</math> and <math>3</math> then gives
−
50bc = 2 * 100/b * 100/c,
+
<math>50bc = 2 \cdot \frac{100}{b} \cdot \frac{100}{c}</math>,
−
from which c = 20/b.
+
from which <math>c = \frac{20}{b}</math>.
−
Comparing columns 1 and 3 gives
+
Comparing columns <math>1</math> and <math>3</math> gives
−
50d*100/c= 2c*100/d,
+
<math>50d \cdot \frac{100}{c}= 2c \cdot \frac{100}{d}</math>,
−
from which d = c/5 = 4/b.
+
from which <math>d = \frac{c}{5} = \frac{4}{b}</math>.
−
Finally, f = 25b, g = 5b, and e = 10. All the entries are positive integers
+
Finally, <math>f = 25b, g = 5b</math>, and <math>e = 10</math>. All the entries are positive integers
−
if and only if b = 1, 2, or 4. The corresponding values for g are 5, 10, and
+
if and only if <math>b = 1, 2,</math> or <math>4</math>. The corresponding values for <math>g</math> are <math>5, 10,</math> and
−
20, and their sum is <math>\boxed{\mathbf{(C)}35} </math>.
+
<math>20</math>, and their sum is <math>\boxed{\mathbf{(C)}35} </math>.
  
 
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.
 
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.

Revision as of 10:12, 20 August 2017

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 (Problems • Answer Key • Resources)
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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12B_Problems/Problem_22&oldid=87132"