Difference between revisions of "2011 AMC 8 Problems/Problem 22"
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Revision as of 21:38, 4 January 2024
Contents
Problem 22 AMC 8 2011
What is the tens digit of 
?
Solution 1
Solution was incorrect insert bruh meme here
Solution 2
Since we want the tens digit, we can find the last two digits of . We can do this by using modular arithmetic.
![\[7^1\equiv 07 \pmod{100}.\]](http://latex.artofproblemsolving.com/d/3/4/d3414d6de021de10a7868a9d21959e6d07811219.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 
.
-Ilovefruits
Solution 3
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 .
-Ilovefruits
Video Solution by OmegaLearn
https://youtu.be/7an5wU9Q5hk?t=1710
Video Solution
https://youtu.be/lxtYmUzQQ8w ~David
Video Solution by WhyMath
~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. 
