Difference between revisions of "2011 AMC 8 Problems/Problem 22"

(Problem)
(Problem)
Line 1: Line 1:
 
==Problem==
 
==Problem==
What is the ones digit of <math>7^{2011}</math>?
+
What is the tens digit of <math>7^{2011}</math>?
  
<math> \textbf{(A) }9\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }7 </math>
+
<math> \textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }7 </math>
  
 
==Video Solution==
 
==Video Solution==

Revision as of 00:48, 4 November 2020

Problem

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

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

Video Solution

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

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

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