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

m (Solution 1)
(Solution 1)
Line 3: Line 3:
  
 
<math> \textbf{(A) }0\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==
 +
https://youtu.be/7an5wU9Q5hk?t=1710
  
 
==Solution 1==
 
==Solution 1==

Revision as of 19:15, 27 October 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 $\boxed{\textbf{(D)}\ 4}$

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