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

m (Solution)
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Let's make a table.
 
Let's make a table.
  
<math>\begin{tabular}[t]{|c|c|c|}
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<cmath>\begin{tabular}[t]{|c|c|c|}
 
\multicolumn{3}{c}{Table}\\hline
 
\multicolumn{3}{c}{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{tabular}</cmath>
  
<center><math>x+19+96=x+1+cc=19+961=114,114+96+a=x+1+114a=x95</math></center>
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<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{tabular}[t]{|c|c|c|}
 
\multicolumn{3}{c}{Table in progress}\\hline
 
\multicolumn{3}{c}{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{tabular}</cmath>
  
<center><math>19+x95+d=x+d76=115+xd=191,114+191+e=x+115e=x190</math></center>
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<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*}
 +
</cmath>
 +
<cmath>\begin{tabular}[t]{|c|c|c|}
 
\multicolumn{3}{c}{Table in progress}\\hline
 
\multicolumn{3}{c}{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{tabular}</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>

Revision as of 17:46, 10 March 2015

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.

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

\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*}

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

\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*} \[\begin{tabular}[t]{|c|c|c|} \multicolumn{3}{c}{Table in progress}\\\hline x&19&96\\\hline 1&x-95&b\\\hline 114&191&x-190\\\hline \end{tabular}\]

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

See also

1996 AIME (ProblemsAnswer KeyResources)
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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