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

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<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>
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==Solution 1==
 
==Solution 1==
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We want the tens digit
 
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So, we take <math>7^{2011}\ (\text{mod }100)</math>. That is congruent to <math>7^{11}\ (\text{mod }100)</math>. From here, it is simple, 7, 49, 43, 01, 07, 49, 43, 01, 07, 49, 43. So the answer is <math>\boxed{\textbf{(D)}\ 4}</math>
 
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==Solution 2==
 
 
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.
 
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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<cmath>7\equiv 07 \pmod{100}.</cmath>
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<cmath>7^1\equiv 07 \pmod{100}.</cmath>
 
<cmath>7^2\equiv 49 \pmod{100}.</cmath>
 
<cmath>7^2\equiv 49 \pmod{100}.</cmath>
 
<cmath>7^3\equiv 43 \pmod{100}.</cmath>
 
<cmath>7^3\equiv 43 \pmod{100}.</cmath>
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-Ilovefruits
 
-Ilovefruits
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==Solution 3==
 
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We can use patterns to figure out the answer.
 
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7 to the power of 1 has no tens digit, so we can ignore it.
 
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7 to the power of 2 is 49.
 
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7 to the power of 3 is 343.
 
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7 to the power of 4 is 2401.
 
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7 to the power of 5 is 16807.
 
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By now, we can notice the pattern. The tens digit for 7 to the power of 1 is 0, then 4, then 4, then 0, then 0. It keeps rotating, 2 fours, and 2 zeros in 4 numbers. If we round up for 2011/4, we get 503. 503 * 4 is 2012. So 7 to the 2012 power has a tens digit of 0, since 2012 is a mutiple of 4, and 7 to the power of 4 has a tens digit of 0. We have to subtract a power from 7 to the 2012 power, so the tens digit goes back from 0 to 4 because if we subtract a power from 7 to the power of 4, we have 7 to the power of 3, which has a tens digit of 4. Hence the answer is <math>\boxed{\textbf{(D)}\ 4}</math>.
 
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-Me
 
  
 
==Video Solution by OmegaLearn==
 
==Video Solution by OmegaLearn==

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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