Difference between revisions of "1990 AIME Problems/Problem 2"
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== Solution == | == Solution == | ||
| − | {{ | + | Suppose that <math>52+6\sqrt{43}</math> is in the form of <math>(a + b\sqrt{43})^2</math>. [[FOIL]]ing yields that <math>52 + 6\sqrt{43} = a^2 + 43b^2 + 2ab\sqrt{43}</math>. This implies that <math>a</math> and <math>b</math> equals one of <math>\pm1, \pm3</math>. The possible [[set]]s are <math>(3,1)</math> and <math>(-3,-1)</math>; the latter can be discarded since the square root must be positive. This means that <math>52 + 6\sqrt{43} = (\sqrt{43} + 3)^2</math>. Repeating this for <math>52-6\sqrt{43}</math>, the only feasible possibility is <math>(\sqrt{43} - 3)^2</math>. |
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| + | Rewriting, we get <math>(\sqrt{43} + 3)^3 - (\sqrt{43} - 3)^3</math>. Using the difference of [[cube]]s, we get that <math>[\sqrt{43} + 3\ - \sqrt{43} + 3]\ [(43 + 6\sqrt{43} + 9) + (43 - 9) + (43 - 6\sqrt{43} + 9)] = (6)(3 \cdot 43 + 9) = 828</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=1990|num-b=1|num-a=3}} | {{AIME box|year=1990|num-b=1|num-a=3}} | ||
Revision as of 19:20, 2 March 2007
Problem
Find the value of 
.
Solution
Suppose that 
is in the form of 
. FOILing yields that 
. This implies that 
and 
equals one of 
. The possible sets are 
and 
; the latter can be discarded since the square root must be positive. This means that 
. Repeating this for 
, the only feasible possibility is 
.
Rewriting, we get 
. Using the difference of cubes, we get that ![$[\sqrt{43} + 3\ - \sqrt{43} + 3]\ [(43 + 6\sqrt{43} + 9) + (43 - 9) + (43 - 6\sqrt{43} + 9)] = (6)(3 \cdot 43 + 9) = 828$](http://latex.artofproblemsolving.com/6/6/c/66cdce57f32b952c40436d122e3c92df3b7dc6c1.png)
.
See also
| 1990 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
