Difference between revisions of "1996 AIME Problems/Problem 1"
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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> | ||
Line 20: | Line 20: | ||
</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> | ||
+ | |||
+ | ==Solution 2== | ||
+ | Use the table from above. Obviously <math>c = 114</math>. Hence <math>a+e = 115</math>. Similarly, <math>1+a = 96 + e \Rightarrow a = 95+e</math>. | ||
+ | |||
+ | Substitute that into the first to get <math>2e = 20 \Rightarrow e=10</math>, so <math>a=105</math>, and so the value of <math>x</math> is just <math>115+x = 210 + 105 \Rightarrow x = \boxed{200}</math> | ||
+ | |||
+ | ==Solution 3== | ||
+ | <cmath>\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}</cmath> | ||
+ | The formula <cmath>e=\frac{1+19}{2}</cmath> can be used. Therefore, <math>e=10</math>. Similarly, <cmath>96=\frac{1+d}{2}</cmath> | ||
+ | So <math>d=191</math>. | ||
+ | |||
+ | Now we have this table: | ||
+ | <cmath>\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}</cmath> | ||
+ | By property of magic squares, observe that | ||
+ | <cmath>x+a+10=19+a+191</cmath> | ||
+ | The <math>a</math>'s cancel! We now have | ||
+ | <cmath>x+10=19+191</cmath> | ||
+ | Thus <math>x=\boxed{200}.</math> | ||
== See also == | == See also == |
Latest revision as of 11:28, 4 August 2021
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 .
![AIME 1996 Problem 01.png](https://wiki-images.artofproblemsolving.com//6/68/AIME_1996_Problem_01.png)
Solution
Let's make a table.
Solution 2
Use the table from above. Obviously . Hence
. Similarly,
.
Substitute that into the first to get , so
, and so the value of
is just
Solution 3
The formula
can be used. Therefore,
. Similarly,
So
.
Now we have this table:
By property of magic squares, observe that
The
's cancel! We now have
Thus
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.