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

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== Solution ==
 
== Solution ==
{{solution}}
+
Let's make a table.
 +
 
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<math>\begin{tabular}[t]{|c|c|c|}
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\multicolumn{3}{c}{Table}\\hline
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x&19&96\\hline
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1&a&b\\hline
 +
c&d&e\\hline
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\end{tabular}</math>
 +
 
 +
<math>x+19+96=x+1+c\Rightarrow c=19+96-1=114</math>.
 +
 
 +
<math>114+96+a=x+1+114\Rightarrow a=x-95</math>
 +
 
 +
<math>\begin{tabular}[t]{|c|c|c|}
 +
\multicolumn{3}{c}{Table in progress}\\hline
 +
x&19&96\\hline
 +
1&x-95&b\\hline
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114&d&e\\hline
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\end{tabular}</math>
 +
 
 +
<math>19+x-95+d=x+d-76=115+x\Rightarrow  d=191</math>.
 +
 
 +
<math>114+191+e=x+115\Rightarrow e=x-190</math>
 +
 
 +
<math>3x-285=x+115\Rightarrow 2x=400\Rightarrow x=\boxed{200}</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1996|before=First Problem|num-a=2}}
 
{{AIME box|year=1996|before=First Problem|num-a=2}}

Revision as of 10:48, 28 February 2008

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.

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

$x+19+96=x+1+c\Rightarrow c=19+96-1=114$.

$114+96+a=x+1+114\Rightarrow a=x-95$

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

$19+x-95+d=x+d-76=115+x\Rightarrow  d=191$.

$114+191+e=x+115\Rightarrow e=x-190$

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

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