Difference between revisions of "2000 AMC 12 Problems/Problem 2"
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We see that <math>a(a^{2000})=a^{2001}.</math> Only answer choice <math>\boxed{\textbf{(A)}}</math> satisfies this requirement. | We see that <math>a(a^{2000})=a^{2001}.</math> Only answer choice <math>\boxed{\textbf{(A)}}</math> satisfies this requirement. | ||
| + | |||
| + | -SirAppel | ||
== Video Solution (Daily Dose of Math) == | == Video Solution (Daily Dose of Math) == | ||
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~Thesmartgreekmathdude | ~Thesmartgreekmathdude | ||
| − | |||
== See also == | == See also == | ||
{{AMC12 box|year=2000|num-b=1|num-a=3}} | {{AMC12 box|year=2000|num-b=1|num-a=3}} | ||
Latest revision as of 16:24, 14 July 2024
- The following problem is from both the 2000 AMC 12 #2 and 2000 AMC 10 #2, so both problems redirect to this page.
Problem
Solution
We can use an elementary exponents rule to solve our problem.
We know that 
. Hence,
Solution edited by armang32324 and integralarefun
Solution 2
We see that 
Only answer choice 
satisfies this requirement.
-SirAppel
Video Solution (Daily Dose of Math)
https://www.youtube.com/watch?v=h0QtF9J0oPs
~Thesmartgreekmathdude
See also
| 2000 AMC 12 (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 AMC 12 Problems and Solutions | |
| 2000 AMC 10 (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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
