2011 AMC 10B Problems/Problem 23
Problem
What is the hundreds digit of 
Solution 1
Since 
we know that 
To compute this, we use a clever application of the binomial theorem.
In all of the other terms, the power of 
is greater than 
and so is equivalent to 
modulo 
which means we can ignore it. We have:
Therefore, the hundreds digit is 
Sidenote: By Euler's Totient Theorem, 
for any 
, so 
and 
. We can then proceed using the clever application of the Binomial Theorem.
Solution 2
We need to compute 
By the Chinese Remainder Theorem, it suffices to compute 
and 
In modulo 
we have 
by Euler's Theorem, and also 
so we have ![\[2011^{2011} = (2011^4)^{502} \cdot 2011^3 \equiv 1^{502} \cdot 3^3 \equiv 3 \pmod{8}.\]](http://latex.artofproblemsolving.com/0/0/0/0005cb6d3a60d8bfe27bc0ad83d3686a04cb0ff2.png)
In modulo 
we have 
by Euler's Theorem, and also 
Therefore, we have 
After finding the solution 
we conclude it is the only one by the Chinese Remainder Theorem. Thus, the hundreds digit is
Solution 3
Notice that the hundreds digit of 2011^2011 won't be affected by 2000. Essentially we could solve the problem by finding the hundreds digit of 11^2011.
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 | ||
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