2011 AMC 10B Problems/Problem 23: Difference between revisions
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Therefore, the hundreds digit is <math>\boxed{\textbf{(D) } 6}</math> | Therefore, the hundreds digit is <math>\boxed{\textbf{(D) } 6}</math> | ||
== See Also== | |||
{{AMC10 box|year=2011|ab=B|num-b=22|num-a=24}} | |||
Revision as of 18:52, 4 June 2011
Problem
What is the hundreds digit of 
?
Solution
Since 
we know 
To compute this, write it as 
and use the binomial theorem.
![\[1^{2011} + 2011 \cdot 1^{2010}10^1 + \frac{2011 \cdot 2010}{2} 1^{2009}10^{2} + \cdots\]](http://latex.artofproblemsolving.com/e/c/8/ec8439c5447c6f903af5ddb05027abe7b27c1f0f.png)
From then on the power of 
is greater than 
and cancel out with 
Therefore, the hundreds digit is 
See Also
| 2011 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 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 | ||
