1996 AIME Problems/Problem 1


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


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 (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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