Difference between revisions of "2011 AMC 10B Problems/Problem 23"
CakeIsEaten (talk | contribs) (→Solution) |
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modulus method | modulus method | ||
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(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). | ||
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pascal's triangle method | pascal's triangle method | ||
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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 02:01, 28 February 2011
Problem
What is the hundreds digit of 
?
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).
