Difference between revisions of "2009 AIME I Problems/Problem 14"
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== See also == | == See also == | ||
{{AIME box|year=2009|n=I|num-b=13|num-a=15}} | {{AIME box|year=2009|n=I|num-b=13|num-a=15}} | ||
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| + | [[Category:Intermediate Algebra Problems]] | ||
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Latest revision as of 13:12, 16 August 2020
Problem
For 
, define 
, where 
. If 
and 
, find the minimum possible value for 
.
Solution
Because the order of the 
's doesn't matter, we simply need to find the number of 
s 
s 
s and 
s that minimize . So let 
and 
represent the number of s, s, s, and s respectively. Then we can write three equations based on these variables. Since there are a total of 
s, we know that 
. We also know that 
and 
. We can now solve these down to two variables:
![\[w = 350 - x - y - z\]](http://latex.artofproblemsolving.com/2/e/f/2efaf625d834188161eb8a033fba648bb0286a80.png)
Substituting this into the second and third equations, we get
![\[x + 2y + 3z = 163\]](http://latex.artofproblemsolving.com/3/b/e/3be0445083f31b8826ca1c819f412b6f54e00edd.png)
and
![\[15x + 80y + 255z = 4395.\]](http://latex.artofproblemsolving.com/d/1/7/d1712bbcb227a1dd16ff28d3dc7f3e3db6258e50.png)
The second of these can be reduced to
![\[3x + 16y + 51z = 879.\]](http://latex.artofproblemsolving.com/b/a/a/baa1f5c639194187ddad1036d6b4bb3d4ada662b.png)
Now we substitute 
from the first new equation into the other new equation.
![\[x = 163 - 2y - 3z\]](http://latex.artofproblemsolving.com/0/2/3/023202aff61bf26fad87e3cfba65d67327eb934d.png)
![\[3(163 - 2y - 3z) + 16y + 51z = 879\]](http://latex.artofproblemsolving.com/b/8/f/b8f72cfe7798221c814ad09dbd829b75c55fc731.png)
![\[489 + 10y + 42z = 879\]](http://latex.artofproblemsolving.com/3/0/4/30478cb0ed3a311ffd40ef087fea203d48f1382b.png)
![\[5y + 21z = 195\]](http://latex.artofproblemsolving.com/e/6/0/e607b9dbd6db04ee38bfc295f50ddcbdc88908f2.png)
Since 
and are integers, the two solutions to this are 
or 
.
If you plug both these solutions in to it is apparent that the second one returns a smaller value. It turns out that 
, 
, 
, and 
, so 
.
See also
| 2009 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
