Difference between revisions of "1996 AIME Problems/Problem 1"
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<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> | ||
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+ | ==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> | ||
== See also == | == See also == |
Revision as of 13:35, 25 July 2019
Contents
[hide]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.
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
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.