Difference between revisions of "1989 AJHSME Problems/Problem 8"
(→Solution 3(Bash)) |
(→Solution 3(Bash)) |
||
| (One intermediate revision by the same user not shown) | |||
| Line 12: | Line 12: | ||
==Solution 3(Bash)== | ==Solution 3(Bash)== | ||
| − | We can just bash it out, getting <math>24(\frac12+\frac13+\frac14)= 12 + 8 + 6 = 26 \ | + | We can just bash it out, getting <math>24(\frac12+\frac13+\frac14)= 12 + 8 + 6 = 26 \Longrightarrow \boxed{\text{E}}</math> |
-fn106068 | -fn106068 | ||
Latest revision as of 10:15, 28 August 2021
Problem
Solution 1
We use the distributive property to get ![\[3\times 4+2\times 4+2\times 3 = 26 \rightarrow \boxed{\text{E}}\]](http://latex.artofproblemsolving.com/6/9/6/696057855d9d7437bb8c492ee09749c3a843647f.png)
Solution 2
Since 
, we have ![\[(2\times 3\times 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) > 2\times 3\times 4 \times 1 = 24\]](http://latex.artofproblemsolving.com/7/3/f/73fb09048546749e763867c37a9daf9b86e9a9df.png)
The only answer choice greater than 
is 
.
Solution 3(Bash)
We can just bash it out, getting 
-fn106068
See Also
| 1989 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
