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

(Solution 2)
(Solution 2)
Line 15: Line 15:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
x_2-x_1&=6</cmath>
 
x_2-x_1&=6</cmath>
<cmath>x_3-x_17&=18</cmath>
+
<cmath>x_3-x_17&=18</cmath>\\
<cmath>x_4-x_1&=42</cmath>
+
<cmath>x_4-x_1&=42</cmath>\\
 
<cmath>x_5-x_1&=90</cmath>
 
<cmath>x_5-x_1&=90</cmath>
 
\end{align*}<math></math>
 
\end{align*}<math></math>

Revision as of 15:10, 22 August 2019

Problem

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

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

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 $3\cdot17 + 2\cdot65 = \boxed{181}$.

Solution 2

Subtracting the first equation from every one of the other equations yields

\begin{align*}
x_2-x_1&=6 (Error compiling LaTeX. Unknown error_msg)
\[x_3-x_17&=18\] (Error compiling LaTeX. Unknown error_msg)

\\

\[x_4-x_1&=42\] (Error compiling LaTeX. Unknown error_msg)

\\

\[x_5-x_1&=90\] (Error compiling LaTeX. Unknown error_msg)

\end{align*}$$ (Error compiling LaTeX. Unknown error_msg)

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

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