2011 AMC 10B Problems/Problem 23

Problem

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

$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) }5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9$

Solution 1

Since $2011 \equiv 11 \pmod{1000},$ we know that $2011^{2011} \equiv 11^{2011} \pmod{1000}.$

To compute this, we use a clever application of the binomial theorem.

$\begin{aligned} 11^{2011} &= (1+10)^{2011} \\ &= 1 + \dbinom{2011}{1} \cdot 10 + \dbinom{2011}{2} \cdot 10^2 + \cdots \end{aligned}$

In all of the other terms, the power of $10$ is greater than $3$ and so is equivalent to $0$ modulo $1000,$ which means we can ignore it. We have:

$\begin{aligned}11^{2011} &\equiv 1 + 2011\cdot 10 + \dfrac{2011 \cdot 2010}{2} \cdot 100 \\ &\equiv 1+20110 + \dfrac{11\cdot 10}{2} \cdot 100\\ &= 1 + 20110 + 5500\\ &\equiv 1 + 110 + 500\\&=611 \pmod{1000} \end{aligned}$

Therefore, the hundreds digit is $\boxed{\textbf{(D) } 6}.$

Side note: By Euler's Totient Theorem, $a^{\phi (1000)} \equiv 1 \pmod{1000}$ for any $a$ relatively prime with 1000, so $a^{400} \equiv 1 \pmod{1000}$ and $11^{2011} \equiv 11^{11} \pmod{1000}$ . We can then proceed using the clever application of the Binomial Theorem.

Solution 2

We need to compute $2011^{2011} \pmod{1000}.$ By the Chinese Remainder Theorem, it suffices to compute $2011^{2011} \pmod{8}$ and $2011^{2011} \pmod{125}.$

In modulo $8,$ we have $2011^4 \equiv 1 \pmod{8}$ by Euler's Theorem, and also $2011 \equiv 3 \pmod{8},$ so we have $2011^{2011} = (2011^4)^{502} \cdot 2011^3 \equiv 1^{502} \cdot 3^3 \equiv 3 \pmod{8}.$

In modulo $125,$ we have $2011^{100} \equiv 1 \pmod{125}$ by Euler's Theorem, and also $2011 \equiv 11 \pmod{125}.$ Therefore, we have $\begin{aligned} 2011^{2011} &= (2011^{100})^{20} \cdot 2011^{11} \\ &\equiv 1^{20} \cdot 11^{11} \\ &= 121^5 \cdot 11 \\ &= (-4)^5 \cdot 11 = -1024 \cdot 11 \\ &\equiv -24 \cdot 11 = -264 \\ &\equiv 111 \pmod{125}. \end{aligned}$

After finding the solution $2011^{2011} \equiv 611 \pmod{1000},$ we conclude it is the only one by the Chinese Remainder Theorem. Thus, the hundreds digit is $\boxed{\textbf{(D) } 6}.$

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 $11^{2011}$ . Powers of $11$ are special because they can be represented by the Pascal's Triangle. Drawing the triangle, there is a theorem that states the powers of $11$ 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 $1, 5, 10, 10, 5,$ and $1$ . Adding all numbers from right to left, we get $161051$ , which is also $11^5$ . In other words, each number is $10^n$ steps from the right side of the row. The hundreds digit is $0$ . We can do the same for $11^{2011}$ , but we only need to find the $3$ digits from the right. Observing, every $3$ number from the right is $1 + 2 + 3... + n$ . So to find the third number from the right on the row of $11^{2011}$ , $f(11^n) = 1 + 2 + 3... + (n-1),$ or $\frac{(2010 \cdot 2011)}{2}$ , or $2021055$ . 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 $2011$ . We must carry the $1$ in $2011$ 's tens digit to the $5$ in $2021055$ 's unit digit to get $\boxed{\textbf{(D) } 6}$ . 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 $2011^2$ has the same last 3 digits as $11^2$ , we can find $11^{2011}$ 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 $11^{51}$ 's last 2 digits are -11, the same as $11^1$ .

From this information, we can figure out $11^{11}$ and $11^{61}$ end in 611. Adding various multiples of 50 to the exponent gives us the fact that $11^{2011}$ 's last digits are 611. We get $\boxed{\textbf{(D) } 6}$ . -ThisUsernameIsTaken

Solution 5 (modular bash)

We are looking for $2011^{2011}\mod\left(1000\right)$ Rewrite as $2011^{\left(1+2+8+16+64+128+256+512+1024\right)}=\left(2011^{1}\cdot2011^{2}\cdot2011^{8}\cdot...\cdot2011^{512}\cdot2011^{1024}\right)$ and we are looking for the $\mod 1000$ of this, which is $(2011\mod1000\cdot2011^{2}\mod1000\cdot...\cdot2011^{1024}\mod1000)\mod1000$ Ignore the digits past the hundreds place and bash it out. $2011\mod1000 = 11$ $2011^{2}\mod1000 =121$ $2011^{4}\mod1000 =641$ $2011^{8}\mod1000 =881$ $2011^{16}\mod1000 =161$ $2011^{32}\mod1000 =921$ $2011^{64}\mod1000 =241$ $2011^{128}\mod1000 =081$ $2011^{256}\mod1000 =561$ $2011^{512}\mod1000 =721$ $2011^{1024}\mod1000 =841$ We do this by substituting each result of the previous power of 2 and multiplying in $\mod 1000$ . Next, we use the expression $(2011\mod1000\cdot2011^{2}\mod1000\cdot...\cdot2011^{1024}\mod1000)\mod1000$ and substitute the values for the different powers in and simplify to $\mod1000$ as we go (by ignoring the thousands digits and beyond during multiplication) to get $2011^{2011}\mod1000=611$ , and the hundreds digit is $\boxed{\textbf{(D) } 6}$ .

~JH. L

Solution 6 (Super Easy)

We know that the hundreds digit of $2011^{11}$ is just the hundreds digit of $11^{11}$ (mod 100). If we actually take a look at the powers of 11, we notice a pattern:

$11^{1} = 11$

$11^{2} = 121$

$11^{3} = 1331$

$11^{4} = 14641$

We notice that the hundreds digit just increases by 1 starting from 1(the hundreds digit of $11^{1}$ ). 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 $11^{11}$ .

$1 + 2 + 3 + 4.... + 11 = 66$ . Thus the answer is $\boxed{\textbf{(D) } 6}$ .

-TheOfficalPro -Ilovebender I believe this solution is wrong, as I can see that the hundreds digit doesn't increase by 1, it goes from 0 to 1 to 3 to 6. A more accurate solution would be the hundreds digit are triangular numbers.

Video Solution by TheBeautyofMath

https://youtu.be/7rosJnqLhs8?t=323

~IceMatrix

2011 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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