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

Revision as of 02:13, 1 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

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

2011 AMC 8 (Problems • Answer Key • Resources)
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.

@@ Line 20: / Line 20: @@
 <cmath>7^{2011}\equiv (7^4)^{502}\times 7^3\equiv 7^3\equiv 343\equiv 43\pmod{100}.</cmath>
 From the above, we can conclude that the last two digits of <math>7^{2011}</math> are 43.  Since they have asked us to find the tens digit, our answer is <math>\boxed{\textbf{(D)}\ 4}</math>.
+~marfu2007
 ==See Also==
 {{AMC8 box|year=2011|num-b=21|num-a=23}}
 {{MAA Notice}}

Page

Toolbox

Search