Difference between revisions of "2022 AMC 10A Problems/Problem 11"
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m-6 &= 1-\frac{12}{m}. | m-6 &= 1-\frac{12}{m}. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
| − | We multiply both sides by <math>m</math>, then rearrange | + | We multiply both sides by <math>m</math>, then rearrange as <cmath>m^2-7m+12=0.</cmath> |
| − | + | By Vieta's Formulas, the sum of such values of <math>m</math> is <math>\boxed{\textbf{(C) } 7}.</math> | |
| − | + | Note that <math>m=3</math> or <math>m=4</math> from the above equation. | |
| − | + | ~MRENTHUSIASM ~KingRavi | |
| − | ~KingRavi | ||
== See Also == | == See Also == | ||
Revision as of 02:28, 12 November 2022
Problem
Ted mistakenly wrote 
as ![$2\cdot\sqrt[m]{\frac{1}{4096}}.$](http://latex.artofproblemsolving.com/5/d/9/5d95b42d9b2ee3a538690f497e19b91b93ed49ec.png)
What is the sum of all real numbers 
for which these two expressions have the same value?
Solution
We are given that ![\[2^m\cdot\sqrt{\frac{1}{4096}} = 2\cdot\sqrt[m]{\frac{1}{4096}}.\]](http://latex.artofproblemsolving.com/6/9/c/69c2cbd459c37ee97c02bdebda6d4ce6f683286d.png)
Converting everything into powers of 
we have
We multiply both sides by , then rearrange as ![\[m^2-7m+12=0.\]](http://latex.artofproblemsolving.com/b/a/7/ba7288683f6d7ca5d6bd5e4ca6374dcd51a462c3.png)
By Vieta's Formulas, the sum of such values of is 
Note that 
or 
from the above equation.
~MRENTHUSIASM ~KingRavi
See Also
| 2022 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
