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Difference between revisions of "1996 AIME Problems/Problem 1"

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Let's make a table.
 
Let's make a table.
  
−
<math>\begin{tabular}[t]{|c|c|c|}
+
<cmath>\begin{array}{|c|c|c|}
−
\multicolumn{3}{c}{Table}\\\hline
+
\multicolumn{3}{c}{\text{Table}}\\\hline
 
x&19&96\\\hline
 
x&19&96\\\hline
 
1&a&b\\\hline
 
1&a&b\\\hline
 
c&d&e\\\hline
 
c&d&e\\\hline
−
\end{tabular}</math>
+
\end{array}</cmath>
  
−
<center><math>\begin{eqnarray*}x+19+96=x+1+c\Rightarrow c=19+96-1=114,\\ 114+96+a=x+1+114\Rightarrow a=x-95\end{eqnarray*}</math></center>
+
<cmath>
 +
\begin{eqnarray*}
 +
x+19+96=x+1+c\Rightarrow c=19+96-1=114,\\ 114+96+a=x+1+114\Rightarrow a=x-95
 +
\end{eqnarray*}
 +
</cmath>
  
−
<math>\begin{tabular}[t]{|c|c|c|}
+
<cmath>\begin{array}{|c|c|c|}
−
\multicolumn{3}{c}{Table in progress}\\\hline
+
\multicolumn{3}{c}{\text{Table in progress}}\\\hline
 
x&19&96\\\hline
 
x&19&96\\\hline
 
1&x-95&b\\\hline
 
1&x-95&b\\\hline
 
114&d&e\\\hline
 
114&d&e\\\hline
−
\end{tabular}</math>
+
\end{array}</cmath>
  
−
<center><math>\begin{eqnarray*}19+x-95+d=x+d-76=115+x\Rightarrow  d=191,\\ 114+191+e=x+115\Rightarrow e=x-190\end{eqnarray*}</math></center>
+
<cmath>
−
 
+
\begin{eqnarray*}19+x-95+d=x+d-76=115+x\Rightarrow  d=191,\\ 114+191+e=x+115\Rightarrow e=x-190
−
<math>\begin{tabular}[t]{|c|c|c|}
+
\end{eqnarray*}
−
\multicolumn{3}{c}{Table in progress}\\\hline
+
</cmath>
 +
<cmath>\begin{array}{|c|c|c|}
 +
\multicolumn{3}{c}{\text{Table in progress}}\\\hline
 
x&19&96\\\hline
 
x&19&96\\\hline
 
1&x-95&b\\\hline
 
1&x-95&b\\\hline
 
114&191&x-190\\\hline
 
114&191&x-190\\\hline
−
\end{tabular}</math>
+
\end{array}</cmath>
  
 
<cmath>3x-285=x+115\Rightarrow 2x=400\Rightarrow x=\boxed{200}</cmath>
 
<cmath>3x-285=x+115\Rightarrow 2x=400\Rightarrow x=\boxed{200}</cmath>
 +
 +
==Solution 2==
 +
Use the table from above. Obviously <math>c = 114</math>. Hence <math>a+e = 115</math>. Similarly, <math>1+a = 96 + e \Rightarrow a = 95+e</math>.
 +
 +
Substitute that into the first to get <math>2e = 20 \Rightarrow e=10</math>, so <math>a=105</math>, and so the value of <math>x</math> is just <math>115+x = 210 + 105 \Rightarrow x = \boxed{200}</math>
 +
 +
==Solution 3==
 +
<cmath>\begin{array}{|c|c|c|}
 +
\multicolumn{3}{c}{\text{Table}}\\\hline
 +
x&19&96\\\hline
 +
1&a&b\\\hline
 +
c&d&e\\\hline
 +
\end{array}</cmath>
 +
The formula <cmath>e=\frac{1+19}{2}</cmath> can be used. Therefore, <math>e=10</math>. Similarly, <cmath>96=\frac{1+d}{2}</cmath>
 +
So <math>d=191</math>.
 +
 +
Now we have this table:
 +
<cmath>\begin{array}{|c|c|c|}
 +
\multicolumn{3}{c}{\text{Table}}\\\hline
 +
x&19&96\\\hline
 +
1&a&b\\\hline
 +
c&191&10\\\hline
 +
\end{array}</cmath>
 +
By property of magic squares, observe that
 +
<cmath>x+a+10=19+a+191</cmath>
 +
The <math>a</math>'s cancel! We now have
 +
<cmath>x+10=19+191</cmath>
 +
Thus <math>x=\boxed{200}.</math>
  
 
== See also ==
 
== See also ==
Line 38: Line 72:
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 10:28, 4 August 2021

Problem

In a magic square, the sum of the three entries in any row, column, or diagonal is the same value. The figure shows four of the entries of a magic square. Find $x$.

AIME 1996 Problem 01.png

Solution

Let's make a table.

\[\begin{array}{|c|c|c|} \multicolumn{3}{c}{\text{Table}}\\\hline x&19&96\\\hline 1&a&b\\\hline c&d&e\\\hline \end{array}\]

\begin{eqnarray*} x+19+96=x+1+c\Rightarrow c=19+96-1=114,\\ 114+96+a=x+1+114\Rightarrow a=x-95 \end{eqnarray*}

\[\begin{array}{|c|c|c|} \multicolumn{3}{c}{\text{Table in progress}}\\\hline x&19&96\\\hline 1&x-95&b\\\hline 114&d&e\\\hline \end{array}\]

\begin{eqnarray*}19+x-95+d=x+d-76=115+x\Rightarrow d=191,\\ 114+191+e=x+115\Rightarrow e=x-190 \end{eqnarray*} \[\begin{array}{|c|c|c|} \multicolumn{3}{c}{\text{Table in progress}}\\\hline x&19&96\\\hline 1&x-95&b\\\hline 114&191&x-190\\\hline \end{array}\]

\[3x-285=x+115\Rightarrow 2x=400\Rightarrow x=\boxed{200}\]

Solution 2

Use the table from above. Obviously $c = 114$. Hence $a+e = 115$. Similarly, $1+a = 96 + e \Rightarrow a = 95+e$.

Substitute that into the first to get $2e = 20 \Rightarrow e=10$, so $a=105$, and so the value of $x$ is just $115+x = 210 + 105 \Rightarrow x = \boxed{200}$

Solution 3

\[\begin{array}{|c|c|c|} \multicolumn{3}{c}{\text{Table}}\\\hline x&19&96\\\hline 1&a&b\\\hline c&d&e\\\hline \end{array}\] The formula \[e=\frac{1+19}{2}\] can be used. Therefore, $e=10$. Similarly, \[96=\frac{1+d}{2}\] So $d=191$.

Now we have this table: \[\begin{array}{|c|c|c|} \multicolumn{3}{c}{\text{Table}}\\\hline x&19&96\\\hline 1&a&b\\\hline c&191&10\\\hline \end{array}\] By property of magic squares, observe that \[x+a+10=19+a+191\] The $a$'s cancel! We now have \[x+10=19+191\] Thus $x=\boxed{200}.$

See also

1996 AIME (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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