Difference between revisions of "1998 AHSME Problems/Problem 4"
Talkinaway (talk | contribs) (→Solution) |
|||
| Line 15: | Line 15: | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:28, 5 July 2013
Problem
Define ![$[a,b,c]$](http://latex.artofproblemsolving.com/f/b/4/fb4fd5b1926f9772de7a9c2479dcc08a1f279ed5.png)
to mean 
, where 
. What is the value of
![$\left[[60,30,90],[2,1,3],[10,5,15]\right]?$](http://latex.artofproblemsolving.com/d/2/c/d2cf272c97594ab2ba19c593e6899385689f28e6.png)
Solution
Note that ![$[ta,tb,tc] = \frac{ta+tb}{tc} = \frac{t(a+b)}{tc} = \frac{a+b}{c} = [a,b,c]$](http://latex.artofproblemsolving.com/c/e/8/ce8ca90fee4369bbae5459b8d8a30fa6986f1d3d.png)
.
Thus ![$[60,30,90] = [2,1,3] = [10,5,15] = \frac{2+1}{3} = 1$](http://latex.artofproblemsolving.com/7/8/f/78f8c2c9a63b9a15b4faa387f8bdc0c554f6a041.png)
, and ![$[1,1,1] = \frac{1+1}{1} = 2 \Longrightarrow \mathbf{(E)}$](http://latex.artofproblemsolving.com/b/7/c/b7caca0a409964a0551df75dcc64ae7af6e0bee5.png)
.
See also
| 1998 AHSME (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 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
