Difference between revisions of "1996 AIME Problems/Problem 1"
Mathgeek2006 (talk | contribs) m (→Solution) |
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Let's make a table. | Let's make a table. | ||
− | <cmath>\begin{ | + | <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{ | + | \end{array}</cmath> |
<cmath> | <cmath> | ||
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</cmath> | </cmath> | ||
− | <cmath>\begin{ | + | <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{ | + | \end{array}</cmath> |
<cmath> | <cmath> | ||
Line 31: | Line 31: | ||
\end{eqnarray*} | \end{eqnarray*} | ||
</cmath> | </cmath> | ||
− | <cmath>\begin{ | + | <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&191&x-190\\\hline | 114&191&x-190\\\hline | ||
− | \end{ | + | \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> |
Revision as of 10:44, 13 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 .
Solution
Let's make a table.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.