Difference between revisions of "2011 AMC 8 Problems/Problem 22"
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We want the tens digit | We want the tens digit | ||
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 an easy bash, 7, 49, 43, 01, 07, 49, 43, 01, 07, 49, 43. So the answer is <math>\boxed{\textbf{(D)}\ 4}</math> | 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 an easy bash, 7, 49, 43, 01, 07, 49, 43, 01, 07, 49, 43. So the answer is <math>\boxed{\textbf{(D)}\ 4}</math> | ||
| + | |||
| + | ==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. | ||
| + | <cmath>7\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>. | ||
==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}} | ||
Revision as of 02:13, 1 November 2020
Problem
What is the tens digit of 
?
Video Solution
https://youtu.be/7an5wU9Q5hk?t=1710
Solution 1
We want the tens digit
So, we take 
. That is congruent to 
. From here, it is an easy bash, 7, 49, 43, 01, 07, 49, 43, 01, 07, 49, 43. So the answer is 
Solution 2
Since we want the tens digit, we can find the last two digits of . We can do this by using modular arithmetic.
![\[7\equiv 07 \pmod{100}.\]](http://latex.artofproblemsolving.com/2/7/2/272f511e6b6b1cf522ab5db931c725cdbdae935c.png)
![\[7^2\equiv 49 \pmod{100}.\]](http://latex.artofproblemsolving.com/b/2/5/b25cc8f14de391dad5f5095a71454e1aac0169d8.png)
![\[7^3\equiv 43 \pmod{100}.\]](http://latex.artofproblemsolving.com/1/f/2/1f227121a150f2b3b6823ed4b74195e3dfced959.png)
![\[7^4\equiv 01 \pmod{100}.\]](http://latex.artofproblemsolving.com/c/b/b/cbb215ba3e81c1db4c3afa7084526922b7aa6698.png)
We can write as 
. Using this, we can say:
![\[7^{2011}\equiv (7^4)^{502}\times 7^3\equiv 7^3\equiv 343\equiv 43\pmod{100}.\]](http://latex.artofproblemsolving.com/5/f/8/5f813cc0903206245e648b3ea84edc41eaf85a6d.png)
From the above, we can conclude that the last two digits of are 43. Since they have asked us to find the tens digit, our answer is .
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. 
