Difference between revisions of "Mock AIME 4 2006-2007 Problems/Problem 7"
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==Problem== | ==Problem== | ||
Find the remainder when <math>3^{3^{3^3}}</math> is divided by 1000. | Find the remainder when <math>3^{3^{3^3}}</math> is divided by 1000. | ||
+ | |||
==Solution== | ==Solution== | ||
+ | Using the [[Carmichael function]], we have <math>\lambda(1000)=100</math>, so <math>3^{100}=1\pmod{1000}</math>. Therefore, letting <math>N=3^{3^3}</math>, we seek to find an <math>n</math> such that <math>N\equiv n\pmod{100}</math> so that <math>3^N\equiv 3^n\pmod{1000}</math>. | ||
− | {{ | + | Using the Carmichael function again, we have <math>\lambda(100)=20</math>, so <math>N=3^{27}\equiv 3^7\pmod{100}\equiv 87\pmod{100}</math>. Therefore <math>n=87</math>, and so we have the following: |
+ | <cmath> | ||
+ | 3^{3^{3^3}}\equiv 3^{87}\pmod{1000}. | ||
+ | </cmath> | ||
+ | Now, | ||
− | + | <cmath>\begin{align*}3^{87}=(3^{20})^4\cdot 3^7&\equiv 401^4\cdot 187\pmod{1000} \\ | |
+ | &\equiv 601\cdot 187\pmod{1000} \\ | ||
+ | &\equiv \boxed{387}\pmod{1000}. | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | == See also== | ||
*[[Mock AIME 4 2006-2007 Problems/Problem 8| Next Problem]] | *[[Mock AIME 4 2006-2007 Problems/Problem 8| Next Problem]] | ||
*[[Mock AIME 4 2006-2007 Problems/Problem 6| Previous Problem]] | *[[Mock AIME 4 2006-2007 Problems/Problem 6| Previous Problem]] | ||
*[[Mock AIME 4 2006-2007 Problems]] | *[[Mock AIME 4 2006-2007 Problems]] | ||
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+ | [[Category:Intermediate Number Theory Problems]] |
Latest revision as of 23:15, 4 January 2010
Problem
Find the remainder when is divided by 1000.
Solution
Using the Carmichael function, we have , so . Therefore, letting , we seek to find an such that so that .
Using the Carmichael function again, we have , so . Therefore , and so we have the following:
Now,