Difference between revisions of "1998 AJHSME Problems/Problem 1"
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===Solution 2=== | ===Solution 2=== | ||
| − | Note that <math>\ | + | Note that <math>\dfrac{6}{x+1}<\dfrac{6}{x}<\dfrac{6}{x-1}</math> (for <math>x>1</math>) and <math>\dfrac{x}{6}<\dfrac{x+1}{6}</math>. Therefore, we just need to compare <math>\dfrac{6}{x+1}</math> and <math>\dfrac{x}{6}</math>. Plugging in <math>x=7</math>, we get <math>\dfrac{3}{4}</math> and <math>\dfrac{7}{6}</math>, respectively, with <math>\dfrac{3}{4}<\dfrac{7}{6}</math>. Thus, the answer is <math>\boxed{(B) \dfrac{6}{x+1}}</math>. |
~By Leon0168 | ~By Leon0168 | ||
Latest revision as of 09:04, 11 October 2025
Problem
For 
, which of the following is the smallest?
Solutions
Solution 1
Plugging 
in for every answer choice would give
From here, we can see that the smallest is answer choice 
Solution 2
Note that 
(for 
) and 
. Therefore, we just need to compare 
and 
. Plugging in , we get 
and 
, respectively, with 
. Thus, the answer is 
.
~By Leon0168
See also
| 1998 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by First question |
Followed by Problem 2 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
