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

The first couple powers of $7$ are $7, 49, 343, 2401, 16807.$ As you can see, the last two digits cycle after every 4 powers. $7^{1}\ (\text{mod }100) \equiv 7^{5}\ (\text{mod }100) \equiv 7^{2009}\ (\text{mod }100).$ From there, we go two more powers. The last two digits are $43$ so the tens digit is $\boxed{\textbf{(D)}\ 4}$

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.

Page

Toolbox

Search

2011 AMC 8 Problems/Problem 22

Problem

Solution

See Also