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> | ||
− | ==Solution== | + | ==Solution 1== |
− | + | 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== | ||
+ | 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
Contents
Problem
What is the tens digit of ?
Solution 1
Since we want the tens digit, we can find the last two digits of . We can do this by using modular arithmetic. We can write as . Using this, we can say: 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 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 .
-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 |
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