1996 AIME Problems/Problem 1

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 $x$ .

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 $x=\boxed{200}.$

1996 AIME (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
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1996 AIME Problems/Problem 1

Contents

Problem

Solution

Solution 2

Solution 3

See also