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

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<math>\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 62 \qquad \textbf{(E)}\ 136</math>
 
<math>\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 62 \qquad \textbf{(E)}\ 136</math>
  
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== Solution A==
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==Solution 1==
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If the power of a prime <math>p^n</math> other than <math>2,5</math> divides <math>g</math>, then from <math>50 \cdot 2 e = 50dg</math> it follows that <math>p^n|e</math>, but then considering the product of the diagonals, <math>p^{2n} |gec</math> but <math>p^{2n} \nmid 100e</math>, contradiction. So the only prime factors of <math>g</math> are <math>2</math> and <math>5</math>.  
+
All the unknown entries can be expressed in terms of <math>b</math>.
 +
Since <math>100e = beh = ceg = def</math>, it follows that <math>h = \frac{100}{b}, g = \frac{100}{c}</math>,
 +
and <math>f = \frac{100}{d}</math>. Comparing rows <math>1</math> and <math>3</math> then gives
 +
<math>50bc = 2 \cdot \frac{100}{b} \cdot \frac{100}{c}</math>,
 +
from which <math>c = \frac{20}{b}</math>.
 +
Comparing columns <math>1</math> and <math>3</math> gives
 +
<math>50d \cdot \frac{100}{c}= 2c \cdot \frac{100}{d}</math>,
 +
from which <math>d = \frac{c}{5} = \frac{4}{b}</math>.
 +
Finally, <math>f = 25b, g = 5b</math>, and <math>e = 10</math>. All the entries are positive integers
 +
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
 +
<math>20</math>, and their sum is <math>\boxed{\mathbf{(C)}35} </math>.
  
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It suffices now to consider the two magic squares comprised of the powers of <math>2</math> and <math>5</math> 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.  
+
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.
  
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The powers of <math>2</math>:
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==Solution 2==
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<center><math>
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We know because this is a multiplicative magic square that each of the following are equal to each other:
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\begin{tabular}{|c|c|c|} \hline 1 & \textit{b} & \textit{c} \\
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<math>100e=ceg=50dg=beh=2cf=50bc=def=2gh</math>
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\hline \textit{d} & \textit{e} & \textit{f} \\
 
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\hline \textit{g} & \textit{h} & 1 \\
 
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\hline \end{tabular}
 
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</math></center>
 
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So <math>1 + 1 + e = g + e + c \Longrightarrow g = 2 - c</math>, so <math>g = 0,1,2</math>. Indeed, we have the magic squares
 
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<center><math>
 
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\begin{tabular}{|c|c|c|} \hline 1 & 0 & 2 \\
 
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\hline 2 & 1 & 0 \\
 
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\hline 0 & 2 & 1 \\
 
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\hline \end{tabular}, \quad \begin{tabular}{|c|c|c|} \hline 1 & 1 & 1 \\
 
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\hline 1 & 1 & 1 \\
 
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\hline 1 & 1 & 1 \\
 
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\hline \end{tabular}, \quad \begin{tabular}{|c|c|c|} \hline 1 & 2 & 0 \\
 
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\hline 0 & 1 & 2 \\
 
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\hline 2 & 0 & 1 \\
 
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\hline \end{tabular},
 
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</math></center>
 
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The powers of <math>5</math>:
 
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<center><math>
 
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\begin{tabular}{|c|c|c|} \hline 2 & \textit{b} & \textit{c} \\
 
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\hline \textit{d} & \textit{e} & \textit{f} \\
 
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\hline \textit{g} & \textit{h} & 0 \\
 
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\hline \end{tabular}
 
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</math></center>
 
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Again, we get <math>2 + e = g + e + c \Longrightarrow g = 0,1,2</math>. However, if we let <math>g = 2, c = 0</math>, then <math>e = d + e + f \Longrightarrow d = f = 0</math>, which obviously gives us a contradiction, and similarly for <math>g = 0, c = 2</math>. For <math>g = 1</math>, we get
 
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<center><math>
 
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\begin{tabular}{|c|c|c|} \hline 2 & 0 & 1 \\
 
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\hline 0 & 1 & 2 \\
 
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\hline 1 & 2 & 0 \\
 
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\hline \end{tabular}
 
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</math></center>
 
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In conclusion, <math>g</math> can be <math>2^0 \cdot 5^1, 2^1 \cdot 5^1, 2^2 \cdot 5^1</math>, and their sum is <math>\boxed{\mathbf{(C)}35} </math>.
 
  
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==Solution B==
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From this we know that <math>50dg=2hg</math>, thus <math>h=25d</math>.
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All the unknown entries can be expressed in terms of b.
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Thus <math>beh=be(25d)</math> and <math>be(25d)=100e</math>. Thus <math>b=\frac{4}{d}</math>
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Since 100e = beh = ceg = def, it follows that h = 100/b, g = 100/c,
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From this we know that <math>50bc=(50)(\frac{4}{d})(c)=50dg</math>. Thus <math>c=\frac{d^2g}{4}</math>.
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and f = 100/d. Comparing rows 1 and 3 then gives
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Now we know from the very beginning that <math>100e=ceg</math> or <math>100=cg</math> or <math>100=\frac{d^2g}{4}(g)</math> or <math>\frac{d^2g^2}{4}</math>. Rearranging the equation <math>100=\frac{d^2g^2}{4}</math> we have <math> (d^2)(g^2)=400</math> or <math>dg=20</math> due to <math>d</math> and <math>g</math> both being positive. Now that <math>dg=20</math> we find all pairs of positive integers that multiply to <math>20</math>. There is <math>(d,g)= (20,1);(10,2);(5,4);(4,5);(2,10);(1,20)</math>. Now we know that <math>b=\frac{4}{d}</math> and b has to be a positive integer. Thus <math>d</math> can only be <math>1</math>, <math>2</math>, or <math>4</math>. Thus <math>g</math> can only be <math>20</math>,<math>10</math>,or <math>5</math>. Thus sum of <math>20+10+5</math> = <math>35</math>. The answer is <math>\boxed{\mathbf{(C)}35} </math>.
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50bc = 2 * 100/b * 100/c,
 
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from which c = 20/b.
 
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Comparing columns 1 and 3 gives
 
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50d*100/c= 2c*100/d,
 
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from which d = c/5 = 4/b.
 
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Finally, f = 25b, g = 5b, and e = 10. All the entries are positive integers
 
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if and only if b = 1, 2, or 4. The corresponding values for g are 5, 10, and
 
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20, 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.
 
  
 
== See also ==
 
== See also ==

Latest revision as of 23:51, 19 October 2021

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 1

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 2

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

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

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