Difference between revisions of "1986 AIME Problems/Problem 4"

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== Solution ==
 
== Solution ==
{{solution}}
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Adding all five [[equation]]s gives us <math>6(x_1 + x_2 + x_3 + x_4 + x_5) = 6(1 + 2 + 4 + 8 + 16)</math> so <math>x_1 + x_2 + x_3 + x_4 + x_5 = 31</math>.  Subtracting this from the fourth given equation gives <math>x_4 = 17</math> and subtracting it from the fifth given equation gives <math>x_5 = 65</math>, so our answer is <math>3\cdot17 + 2\cdot65 = 181</math>.
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== See also ==
 
== See also ==
* [[1986 AIME Problems]]
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{{AIME box|year=1986|num-b=3|num-a=5}}
  
{{AIME box|year=1986|num-b=3|num-a=5}}
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[[Category:Introductory Algebra Problems]]

Revision as of 13:40, 16 February 2007

Problem

Determine $\displaystyle 3x_4+2x_5$ if $\displaystyle x_1$, $\displaystyle x_2$, $\displaystyle x_3$, $\displaystyle x_4$, and $\displaystyle x_5$ satisfy the system of equations below.

$\displaystyle 2x_1+x_2+x_3+x_4+x_5=6$
$\displaystyle x_1+2x_2+x_3+x_4+x_5=12$
$\displaystyle x_1+x_2+2x_3+x_4+x_5=24$
$\displaystyle x_1+x_2+x_3+2x_4+x_5=48$
$\displaystyle x_1+x_2+x_3+x_4+2x_5=96$

Solution

Adding all five equations gives us $6(x_1 + x_2 + x_3 + x_4 + x_5) = 6(1 + 2 + 4 + 8 + 16)$ so $x_1 + x_2 + x_3 + x_4 + x_5 = 31$. Subtracting this from the fourth given equation gives $x_4 = 17$ and subtracting it from the fifth given equation gives $x_5 = 65$, so our answer is $3\cdot17 + 2\cdot65 = 181$.

See also

1986 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions