2011 AMC 10B Problems/Problem 23
Contents
[hide]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
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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