2011 AMC 8 Problems/Problem 22

Problem

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

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 $\boxed{\textbf{(D)}\ 4}$ . ~marfu2007

Video Solution

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

2011 AMC 8 (Problems • Answer Key • Resources)
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2011 AMC 8 Problems/Problem 22

Contents

Problem

Solution 1

Solution 2

Video Solution

See Also