2011 AMC 10B Problems/Problem 23
Contents
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
Side note: By Euler's Totient Theorem, for any
relatively prime with 1000, 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 won't be affected by
. Essentially we could solve the problem by finding the hundreds digit of
. Powers of
are special because they can be represented by the Pascal's Triangle. Drawing the triangle, there is a theorem that states the powers of
can be found by reading rows of the triangle and adding extra numbers up. [add source] For example, the sixth row of the triangle is
and
. Adding all numbers from right to left, we get
, which is also
. In other words, each number is
steps from the right side of the row. The hundreds digit is
. We can do the same for
, but we only need to find the
digits from the right. Observing, every
number from the right is
. So to find the third number from the right on the row of
,
or
, or
. The last digit is five, but we must remember to add the number on the right of it, which, by observing other rows is obviously
. We must carry the
in
's tens digit to the
in
's unit digit to get
. The one at the very end of the row doesn't affect anything, so we can leave it alone.
-jackshi2006
Solution 4 (naive solution)
Since we are only looking at the last 3 digits and has the same last 3 digits as
, we can find
instead.
After this, we can repeatedly multiply the last 3 digits by 11 and take the last 3 digits of that product. We discover that
's last 2 digits are -11, the same as
.
From this information, we can figure out and
end in 611. Adding various multiples of 50 to the exponent gives us the fact that
's last digits are 611. We get
.
-ThisUsernameIsTaken
Solution 5 (modular bash)
We are looking for Rewrite as
and we are looking for the
of this, which is
Ignore the digits past the hundreds place and bash it out.
We do this by substituting each result of the previous power of 2 and multiplying in
. Next, we use the expression
and substitute the values for the different powers in and simplify to
as we go (by ignoring the thousands digits and beyond during multiplication) to get
, and the hundreds digit is
.
~JH. L
Solution 6 (Super Easy)
We know that the hundreds digit of is just the hundreds digit of
(mod 100). If we actually take a look at the powers of 11, we notice a pattern:
We notice that the hundreds digit just increases by 1 starting from 1(the hundreds digit of ). All we have to do now is add all the numbers from 1 to 11 (since the hundreds digit increases by 1 with every power of 11 that comes after) to get the hundreds digit of
.
. Thus the answer is
.
-TheOfficalPro
Video Solution by TheBeautyofMath
https://youtu.be/7rosJnqLhs8?t=323
~IceMatrix
See Also
2011 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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