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.

-Extremelysupercooldude: In your solution, you find the hundreds digit of $2011^{11}$ , but the problem asks for the hundreds digit of $2011^{2011}$ , but you never cite way the two hundreds digits are the same.

-Wesserwes7254: I think this is a more revised version of this solution: We know that the hundreds digit of $2011^{2011}$ is just the hundreds digit of $11^{11}$ (mod 100). If we actually take a look at the powers of 11 modulo 1000, we notice a pattern:

\[11^{1} &\equiv 011 \Mod{100}\] (Error compiling LaTeX. Unknown error_msg)

\[11^{2} &\equiv 121 \Mod{100}\] (Error compiling LaTeX. Unknown error_msg)

\[11^{3} &\equiv 331 \Mod{100}\] (Error compiling LaTeX. Unknown error_msg)

\[11^{4} &\equiv 641 \Mod{100}\] (Error compiling LaTeX. Unknown error_msg)

\[11^{5} &\equiv 051 \Mod{100}\] (Error compiling LaTeX. Unknown error_msg)

\[11^{6} &\equiv 561 \Mod{100}\] (Error compiling LaTeX. Unknown error_msg)

Notice how the units digit is always one, the tens digit is always the previous tens digit plus the ones digit (or one) and the hundreds digit is always the previous hundreds digit plus the previous tens digit. Knowing this, we confidently repeat this pattern without actually multiply the previous term by 11 out (if you generally a few numbers by 11, you can see why this pattern holds).

\[11^{7} &\equiv 171 \Mod{1000}\] (Error compiling LaTeX. Unknown error_msg)

\[11^{8} &\equiv 881 \Mod{1000}\] (Error compiling LaTeX. Unknown error_msg)

\[11^{9} &\equiv 691 \Mod{1000}\] (Error compiling LaTeX. Unknown error_msg)

\[11^{10} &\equiv 601 \Mod{1000}\] (Error compiling LaTeX. Unknown error_msg)

\[11^{11} &\equiv 611 \Mod{1000}\] (Error compiling LaTeX. Unknown error_msg)

Since $11^{11}$ ends in -11, the pattern "cycles" every 10 times. $\frac{2011}{10}=201 r1$ . $(201 \cdot 2011)}$ (Error compiling LaTeX. Unknown error_msg)\equiv 6 /Mod{10} $, meaning that$ 11^{2011} &\equiv $2011^{2011} &\equiv 611 /Mod{1000}, giving us a hundreds digit of$ (Error compiling LaTeX. Unknown error_msg)\boxed{\textbf{(D) } 6}.$

Video Solution by TheBeautyofMath

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

~IceMatrix

Video Solution by OmegaLearn

https://youtu.be/-H4n-QplQew?t=1423

~ pi_is_3.14

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