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

Since we want the tens digit, we can find the last two digits of $7^{2011}$. We can do this by using modular arithmetic. \[7^1\equiv 07 \pmod{100}.\] \[7^2\equiv 49 \pmod{100}.\] \[7^3\equiv 43 \pmod{100}.\] \[7^4\equiv 01 \pmod{100}.\] 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 2

We can use patterns to figure out the answer. 7 to the power of 2 is 49. So, the tens digit is 4. 7 to the power of 3 is 343. So, the tens digit is 4. 7 to the power of 4 is 2401.So, the tens digit is 0. 7 to the power of 5 is 16807. So, the tens digit is 0. By now, we can notice the pattern. The tens digit for 7 to the power of 2 is 4, then 4, then 0, then 0. It keeps on going, 2 fours, and then 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 $\boxed{\textbf{(D)}\ 4}$.

-Ilovefruits

Video Solution by OmegaLearn

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

Video Solution

https://youtu.be/lxtYmUzQQ8w ~David

Video Solution by WhyMath

https://youtu.be/Jyf_ILTO3nI

~savannahsolver

See Also

2011 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png