Difference between revisions of "2011 AMC 10B Problems/Problem 23"

(Solution)
(Solution)
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modulus method
 
modulus method
 +
  
 
(2000  + 11) ^ 2011 mod 1000 \n
 
(2000  + 11) ^ 2011 mod 1000 \n
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So we know the last three digits of 2011 ^ 2011 is 611, and so the hundreds digit is 6 (D).
 
So we know the last three digits of 2011 ^ 2011 is 611, and so the hundreds digit is 6 (D).
 +
  
 
pascal's triangle method
 
pascal's triangle method
 +
  
 
First we try some multiplications. But we will only look at some of the last digits
 
First we try some multiplications. But we will only look at some of the last digits

Revision as of 01:01, 28 February 2011

Problem

What is the hundreds digit of $2011^{2011}$?

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8$

Solution

modulus method


(2000 + 11) ^ 2011 mod 1000 \n

11^2011 mod 1000

(10 + 1)^2011 mod 1000

2011C2 * 10^2 + 2011C1 * 10 + 1 mod 1000

500 + 110 + 1 mod 1000

611 mod 1000

So we know the last three digits of 2011 ^ 2011 is 611, and so the hundreds digit is 6 (D).


pascal's triangle method


First we try some multiplications. But we will only look at some of the last digits

First, we see that 2011 ^ 2 gives the 3 last digits 121

Then 2011 ^ 3 gives 1331

2011 ^ 4 gives 2641

If we continue, we eventually see that the thousand's term and the hundred's term are part of the triangle numbers, and is part of the pascal triangle.

The hundred's digit is the last digit of the nth triangle number, which in our case is 2011.

Therefore, we just do 2011(2012) / 2 => last digit is 6 (D).