Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (→‎Problem 22 AMC 8 2011)
 
(36 intermediate revisions by 15 users not shown)
Line 5: Line 5:
  
 
==Solution 1==
 
==Solution 1==
−
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>
+
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.
 +
<cmath>7^1\equiv 07 \pmod{100}.</cmath>
 +
<cmath>7^2\equiv 49 \pmod{100}.</cmath>
 +
<cmath>7^3\equiv 43 \pmod{100}.</cmath>
 +
<cmath>7^4\equiv 01 \pmod{100}.</cmath>
 +
We can write <math>7^{2011}</math> as <math>(7^4)^{502}\times 7^3</math>.  Using this, we can say:
 +
<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>.
 +
 
 +
-Ilovefruits
 +
 
 
==Solution 2==
 
==Solution 2==
−
We want the tens digit
+
We can use patterns to figure out the answer.
 +
7 to the power of 2 is 49. So, the tens digit is 4.
 +
7 to the power of 3 is 343. So, the tens digit is 4.
 +
7 to the power of 4 is 2401.So, the tens digit is 0.
 +
7 to the power of 5 is 16807. So, the tens digit is 0.
 +
By now, we can notice the pattern. The tens digit for 7 to the power of 2 is 4, then 4, then 0, then 0. It keeps on going, 2 fours, and then 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>.
 +
 
 +
-Ilovefruits
 +
 
 +
==Video Solution by OmegaLearn==
 +
https://youtu.be/7an5wU9Q5hk?t=1710
 +
 
 +
==Video Solution==
 +
https://youtu.be/lxtYmUzQQ8w  ~David
 +
 
 +
==Video Solution by WhyMath==
 +
https://youtu.be/Jyf_ILTO3nI
 +
 
 +
~savannahsolver
  
 
==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 21:54, 15 January 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

Solution 2

We can use patterns to figure out the answer. 7 to the power of 2 is 49. So, the tens digit is 4. 7 to the power of 3 is 343. So, the tens digit is 4. 7 to the power of 4 is 2401.So, the tens digit is 0. 7 to the power of 5 is 16807. So, the tens digit is 0. By now, we can notice the pattern. The tens digit for 7 to the power of 2 is 4, then 4, then 0, then 0. It keeps on going, 2 fours, and then 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 $\boxed{\textbf{(D)}\ 4}$.

-Ilovefruits

Video Solution by OmegaLearn

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

Video Solution

https://youtu.be/lxtYmUzQQ8w ~David

Video Solution by WhyMath

https://youtu.be/Jyf_ILTO3nI

~savannahsolver

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems/Problem_22&oldid=210889"