Difference between revisions of "1996 AIME Problems/Problem 1"
Line 5: | Line 5: | ||
== Solution == | == Solution == | ||
− | {{ | + | Let's make a table. |
+ | |||
+ | <math>\begin{tabular}[t]{|c|c|c|} | ||
+ | \multicolumn{3}{c}{Table}\\hline | ||
+ | x&19&96\\hline | ||
+ | 1&a&b\\hline | ||
+ | c&d&e\\hline | ||
+ | \end{tabular}</math> | ||
+ | |||
+ | <math>x+19+96=x+1+c\Rightarrow c=19+96-1=114</math>. | ||
+ | |||
+ | <math>114+96+a=x+1+114\Rightarrow a=x-95</math> | ||
+ | |||
+ | <math>\begin{tabular}[t]{|c|c|c|} | ||
+ | \multicolumn{3}{c}{Table in progress}\\hline | ||
+ | x&19&96\\hline | ||
+ | 1&x-95&b\\hline | ||
+ | 114&d&e\\hline | ||
+ | \end{tabular}</math> | ||
+ | |||
+ | <math>19+x-95+d=x+d-76=115+x\Rightarrow d=191</math>. | ||
+ | |||
+ | <math>114+191+e=x+115\Rightarrow e=x-190</math> | ||
+ | |||
+ | <math>3x-285=x+115\Rightarrow 2x=400\Rightarrow x=\boxed{200}</math> | ||
== See also == | == See also == | ||
{{AIME box|year=1996|before=First Problem|num-a=2}} | {{AIME box|year=1996|before=First Problem|num-a=2}} |
Revision as of 10:48, 28 February 2008
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 |