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Difference between revisions of "2011 AMC 8 Problems/Problem 22"
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==Solution 1==
 
==Solution 1==
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The first couple powers of <math>7</math> are <math>7, 49, 343, 2401, 16807.</math> As you can see, the last two digits cycle after every 4 powers. <math>7^{1}\ (\text{mod }100) \equiv 7^{5}\ (\text{mod }100) \equiv 7^{2009}\ (\text{mod }100).</math> From there, we go two more powers. The last two digits are <math>43</math> so the tens digit is <math>\boxed{\textbf{(D)}\ 4}</math>
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Since we want the tens digit, we can find the last two digits of <math>7^{2011}</math>.  We can do this by using modular arithmetic.
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==Solution 2==
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<cmath>7^1\equiv 07 \pmod{100}.</cmath>
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We want the tens digit
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<cmath>7^2\equiv 49 \pmod{100}.</cmath>
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So, we take <math>7^{2009}\ (\text{mod }100)</math>. That is congruent to <math>7^9\ (\text{mod}100)</math>. From here, it is an easy bash, 7, 49, 43, 01, 07, 49, 43, 01, 07
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<cmath>7^3\equiv 43 \pmod{100}.</cmath>
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<cmath>7^4\equiv 01 \pmod{100}.</cmath>
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We can write <math>7^{2011}</math> as <math>(7^4)^{502}\times 7^3</math>.  Using this, we can say:
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<cmath>7^{2011}\equiv (7^4)^{502}\times 7^3\equiv 7^3\equiv 343\equiv 43\pmod{100}.</cmath>
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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>.
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-Ilovefruits
 +
 
 +
==Video Solution by OmegaLearn==
 +
https://youtu.be/7an5wU9Q5hk?t=1710
 +
 
 +
==Video Solution==
 +
https://youtu.be/lxtYmUzQQ8w  ~David
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2011|num-b=21|num-a=23}}
 
{{AMC8 box|year=2011|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 10:34, 15 July 2024

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

Since we want the tens digit, we can find the last two digits of $7^{2011}$. We can do this by using modular arithmetic. \[7^1\equiv 07 \pmod{100}.\] \[7^2\equiv 49 \pmod{100}.\] \[7^3\equiv 43 \pmod{100}.\] \[7^4\equiv 01 \pmod{100}.\] 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 $\boxed{\textbf{(D)}\ 4}$.

-Ilovefruits

Video Solution by OmegaLearn

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

Video Solution

https://youtu.be/lxtYmUzQQ8w ~David

See Also

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

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