Difference between revisions of "1996 AIME Problems/Problem 1"
Mathgeek2006 (talk | contribs) m (→Solution) |
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Line 7: | Line 7: | ||
Let's make a table. | Let's make a table. | ||
− | < | + | <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}</ | + | \end{tabular}</cmath> |
− | < | + | <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> | ||
− | < | + | <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}</ | + | \end{tabular}</cmath> |
− | < | + | <cmath> |
− | + | \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*} |
+ | </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}</ | + | \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 .
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.