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

Latest revision as of 15:25, 17 November 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 $\begin{align*} x_2-x_1&=6\\ x_3-x_1&=18\\ x_4-x_1&=42\\ x_5-x_1&=90 \end{align*}$ Thus $\begin{align*} 2x_1+x_2+x_3+x_4+x_5&=6\\ 2x_1+(x_1+6)+(x_1+18)+(x_1+42)+(x_1+90)&=6\\ 6x_1+156&=6\\ x_1&=-25 \end{align*}$ Using the previous equations, $3x_4+2x_5=3(x_1+42)+2(x_1+90)=\boxed{181}$

~ Nafer

@@ Line 27: / Line 27: @@
 \end{align*}</cmath>
 Using the previous equations,
-<cmath>3x_4+2x_5=3(x_1+42)=2(x_1+90)=\boxed{181}</cmath>
+<cmath>3x_4+2x_5=3(x_1+42)+2(x_1+90)=\boxed{181}</cmath>
 ~ Nafer

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

Latest revision as of 15:25, 17 November 2019

Contents

Problem

Solution

Solution 2

See also