Difference between revisions of "1986 AIME Problems/Problem 4"
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2x_1+(x_1+6)+(x_1+18)+(x_1+42)+(x_1+90)&=6\\ | 2x_1+(x_1+6)+(x_1+18)+(x_1+42)+(x_1+90)&=6\\ | ||
6x_1+156&=6\\ | 6x_1+156&=6\\ | ||
− | x_1=-25 | + | x_1&=-25 |
\end{align*}</cmath> | \end{align*}</cmath> | ||
Using the previous equations, | Using the previous equations, | ||
− | <cmath>3x_4+2x_5=3(x_1+42) | + | <cmath>3x_4+2x_5=3(x_1+42)+2(x_1+90)=\boxed{181}</cmath> |
~ Nafer | ~ Nafer |
Latest revision as of 16:25, 17 November 2019
Contents
Problem
Determine if
,
,
,
, and
satisfy the system of equations below.
![$2x_1+x_2+x_3+x_4+x_5=6$](http://latex.artofproblemsolving.com/6/1/5/6153cca66b47747c7acb479bb4b84465e22bbcff.png)
![$x_1+2x_2+x_3+x_4+x_5=12$](http://latex.artofproblemsolving.com/6/4/6/646b9a02ed66b93600a186f55d1fac2a6e9f935f.png)
![$x_1+x_2+2x_3+x_4+x_5=24$](http://latex.artofproblemsolving.com/9/d/3/9d38eae8f3d3f01e7d7e16b0cd83d7310b624ccf.png)
![$x_1+x_2+x_3+2x_4+x_5=48$](http://latex.artofproblemsolving.com/4/a/0/4a042838b7798d4f7a80a38df7c9d85418553e7a.png)
![$x_1+x_2+x_3+x_4+2x_5=96$](http://latex.artofproblemsolving.com/6/7/0/670b025ed03e65cf8f599f37c2f1da1ba31df0c2.png)
Solution
Adding all five equations gives us so
. Subtracting this from the fourth given equation gives
and subtracting it from the fifth given equation gives
, so our answer is
.
Solution 2
Subtracting the first equation from every one of the other equations yields
Thus
Using the previous equations,
~ Nafer
See also
1986 AIME (Problems • Answer Key • Resources) | ||
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 |
- AIME Problems and Solutions
- American Invitational Mathematics Examination
- Mathematics competition resources
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.