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Difference between revisions of "Mock AIME 4 2006-2007 Problems/Problem 7"

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==Solution==
 
==Solution==
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<math>\phi (1000)=400</math>, so <math>3^{400}=1\bmod 1000</math>.
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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>.
  
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Therefore, we want <math>3^{27} \bmod 400</math>.
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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:
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<cmath>
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3^{3^{3^3}}\equiv 3^{87}\pmod{1000}.
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</cmath>
  
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Since <math>400=2^4*5^2</math>, we want <math>3^{27}\bmod 16</math> and <math>3^{27} \bmod 25</math>.
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Now,
  
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<math>3^{4} \equiv 1\bmod 16</math>, so <math>3^{27}\equiv 11\bmod 16</math>.
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<cmath>\begin{align*}3^{87}=(3^{20})^4\cdot 3^7&\equiv 401^4\cdot 187\pmod{1000} \\
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&\equiv 601\cdot 187\pmod{1000} \\
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Since <math>3^{\phi(25)}=3^{20}\equiv 1\bmod 20</math>, <math>3^{27}\equiv 3^7\equiv 18\bmod 25</math>.
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&\equiv \boxed{387}\pmod{1000}.
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\end{align*}</cmath>
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The only number <math>\bmod 400</math> that is <math>11\bmod 16</math> and <math>18\bmod 25</math> is <math>43\bmod 400</math>. Therefore,
 
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<math>3^{3^{3^3}}\equiv 3^{43} \bmod 1000</math>.
 
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Since <math>1000=8*125</math>, we want to find <math>3^{43}\bmod 8</math> and <math>3^{43}\bmod 125</math>.
 
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Since <math>3^4\equiv 1\bmod 8</math>, <math>3^{43}\equiv 27\equiv 3\bmod 8</math>.
 
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And since <math>3^5\equiv -7\bmod 125</math>, <math>3^{10}\equiv 49\bmod 125</math>, <math>3^{20}\equiv 26\bmod 125</math>, <math>3^{40}\equiv 1\bmod 125</math>.
 
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We have gotten somewhere.
 
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<math>3^{43}\equiv 27\bmod 125</math>
 
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The only number <math>\bmod 1000</math> that satisfies <math>3\bmod 8</math> and <math>27\bmod 125</math> is <math>\boxed{027}\bmod 1000</math>.
 
  
 
== See also==
 
== See also==
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*[[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 00:15, 5 January 2010

Problem

Find the remainder when $3^{3^{3^3}}$ is divided by 1000.

Solution

Using the Carmichael function, we have $\lambda(1000)=100$, so $3^{100}=1\pmod{1000}$. Therefore, letting $N=3^{3^3}$, we seek to find an $n$ such that $N\equiv n\pmod{100}$ so that $3^N\equiv 3^n\pmod{1000}$.

Using the Carmichael function again, we have $\lambda(100)=20$, so $N=3^{27}\equiv 3^7\pmod{100}\equiv 87\pmod{100}$. Therefore $n=87$, and so we have the following: \[3^{3^{3^3}}\equiv 3^{87}\pmod{1000}.\]

Now,

\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*}

See also

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