2011 AMC 8 Problems/Problem 22

Problem 22 AMC 8

What is the tens digit of $7^{2011}$ ?

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }7$

Solution 1

We want to know tens digit So, we take $7^{2011}\ (\text{mod }100)$ . That is congruent to $7^{11}\ (\text{mod }100)$ . From here, it is simple, 7, 49, 43, 01, 07, 49, 43, 01, 07, 49, 43. So the answer is $\boxed{\textbf{(D)}\ 4}$

-bruh made changes

Solution 2

Since we want the tens digit, we can find the last two digits of $7^{2011}$ . We can do this by using modular arithmetic. $7\equiv 07 \pmod{100}.$ $7^2\equiv 49 \pmod{100}.$ $7^3\equiv 43 \pmod{100}.$ $7^4\equiv 01 \pmod{100}.$ We can write $7^{2011}$ as $(7^4)^{502}\times 7^3$ . Using this, we can say: $7^{2011}\equiv (7^4)^{502}\times 7^3\equiv 7^3\equiv 343\equiv 43\pmod{100}.$ From the above, we can conclude that the last two digits of $7^{2011}$ are 43. Since they have asked us to find the tens digit, our answer is $\boxed{\textbf{(D)}\ 4}$ .

-Ilovefruits

Solution 3

We can use patterns to figure out the answer. 7 to the power of 1 has no tens digit, so we can ignore it. 7 to the power of 2 is 49. 7 to the power of 3 is 343. 7 to the power of 4 is 2401. 7 to the power of 5 is 16807. By now, we can notice the pattern. The tens digit for 7 to the power of 1 is 0, then 4, then 4, then 0, then 0. It keeps rotating, 2 fours, and 2 zeros in 4 numbers. If we round up for 2011/4, we get 503. 503 * 4 is 2012. So 7 to the 2012 power has a tens digit of 0, since 2012 is a mutiple of 4, and 7 to the power of 4 has a tens digit of 0. We have to subtract a power from 7 to the 2012 power, so the tens digit goes back from 0 to 4 because if we subtract a power from 7 to the power of 4, we have 7 to the power of 3, which has a tens digit of 4. Hence the answer is $\boxed{\textbf{(D)}\ 4}$ .

-Me

Video Solution by OmegaLearn

https://youtu.be/7an5wU9Q5hk?t=1710

Video Solution

https://youtu.be/lxtYmUzQQ8w ~David

2011 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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