2011 AMC 8 Problems/Problem 22

Revision as of 10:50, 12 November 2019 by Firebolt360 (talk | contribs) (Solution 2)

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

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 $\boxed{\textbf{(D)}\ 4}$

Solution 2

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 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

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