Difference between revisions of "2010 AMC 8 Problems/Problem 2"
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<math>5 @ 10 = \frac{5\times 10}{5+10} = \frac{50}{15} = \frac{10}{3}</math> | <math>5 @ 10 = \frac{5\times 10}{5+10} = \frac{50}{15} = \frac{10}{3}</math> | ||
| − | Thus, answer choice <math>\boxed{\textbf{(D)}\ \frac{10}{3}}</math> is correct | + | Thus, answer choice <math>\boxed{\textbf{(D)}\ \frac{10}{3}}</math> is correct. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2010|num-b=1|num-a=3}} | {{AMC8 box|year=2010|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:51, 2 November 2020
Problem
If 
for 
positive integers, then what is 
?
Solution
Substitute 
and 
into the expression for 
to get:
Thus, answer choice 
is correct.
See Also
| 2010 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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. 
