Difference between revisions of "2000 AMC 12 Problems/Problem 2"
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| + | ==Solution 3== | ||
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| + | Say you somehow forgot the basic rules of exponents, you could just deduce <math>2000^{2000}</math> is a line of 2000 2000's being multiplied, and if we multiplied by this line by an additional 2000, we would then have <math>\boxed{\textbf{(A)}}</math> 2000's lined up multiplying each other. | ||
== Video Solution (Daily Dose of Math) == | == Video Solution (Daily Dose of Math) == | ||
Revision as of 19:36, 27 October 2024
- The following problem is from both the 2000 AMC 12 #2 and 2000 AMC 10 #2, so both problems redirect to this page.
Contents
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
Solution 3
Say you somehow forgot the basic rules of exponents, you could just deduce 
is a line of 2000 2000's being multiplied, and if we multiplied by this line by an additional 2000, we would then have 2000's lined up multiplying each other.
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. 
