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

Revision as of 10:33, 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

Video Solution by WhyMath

https://youtu.be/Jyf_ILTO3nI

~savannahsolver

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.

@@ Line 13: / Line 13: @@
 <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==
-We can use patterns to figure out the answer.
-to the power of 2 is 49. So, the tens digit is 4.
-to the power of 3 is 343. So, the tens digit is 4.
-to the power of 4 is 2401.So, the tens digit is 0.
-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

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