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Find the remainder when $444^{444^{444}}$ is divided by $7$.

My approach :

$E(7) = 6 $

$444^{444} \pmod 6 = 0$

so , $444^0 \pmod 7 = 1$

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    $444^{444^{444}}(mod 7) \neq 444^{444^{444}(mod 7)}(mod 7)$2012-06-27
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    If by E(7) you mean $\varphi(7)=6$, then yeah.2012-06-27
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    @avatar: But OP is doing $\bmod6$ in the exponent, not $\bmod7$. See [Euler's theorem](http://en.wikipedia.org/wiki/Euler's_theorem).2012-06-27
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    Sorry!,i didn't notice that.2012-06-27

3 Answers 3

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Yes your approach is indeed correct. I assume by $E(7)$ you mean the Euler totient function $\phi(7)$.

By Euler's theorem/ Fermat's little theorem, we have $$444^{6} \equiv 1 \pmod{7}$$ Now $6$ divides $444$. Hence, $444^{444} = 6M$. Hence, $$444^{444^{444}} = 444^{6M} = \left( 444^6\right)^M \equiv 1^M \pmod{7} \equiv 1 \pmod{7}$$

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    Yes I meant the same.Thanks sir @marvis.2012-06-27
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$a^{\phi(p)}=1(\mod p)$.Here $444^{444}=0(\mod \phi(7)) \implies 444^{444}=k*\phi(7)$ for some integer k.Then, $444^{444^{444}}(\mod 7)= 444^{k*\phi(7)}(\mod 7) = (444^{\phi(7)})^k(\mod 7)=1^k(\mod 7) = 1 $

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step 1 : Try to bring big no's into a form so that we get remainder of $-1 $ or $1$ , so that it will be easy for us to simplify

$$444 = 4*111 $$ as $ \frac {111}7 $ gives a remainder of $-1$ . So our goal of converting a bigger number to number which gives remainder $1$ or $-1$ is attained and as $$ -1^{even} = 1$$

so $$ 444^x /7 = (4^x * 111^x )/7 = 4^x / 7 $$ $$( 111^x \ divided \ by\ 7\ \ \ gives \ remainder \ of \ -1^x\ which\ is\ equal\ to\ 1\ as\ x\ is\ even\ )$$

Where $$ x = 444^{444} (even) $$ So

$$4^x= 4^{444^{444}} = 2^{888^{444}} $$

step 2 : Observe the pattern of the remainders

$$ 2^0 mod 7 = 1 $$

$$ 2^1 mod 7 = 2 $$

$$ 2^2 mod 7 = 4 $$


$$ 2^3 mod 7 = 1 $$

$$ 2^4 mod 7 = 2 $$

$$2^5 mod 7 = 4 $$


so here, for a period of $ 3 $ remainder $ 1 $ repeats

here $ 888 $ is a multiple of $3$ , so

$$ 2^{888}\ mod \ 7 =\ 1 $$

$$ 1^{444} \ mod \ 7 = \ 1 $$

Hence Answer is $ 1 $

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    For some basic information about writing math at this site see e.g. [here](http://meta.math.stackexchange.com/questions/5020/), [here](http://meta.stackexchange.com/a/70559/155238), [here](http://meta.math.stackexchange.com/questions/1773/) and [here](http://math.stackexchange.com/editing-help#latex).2012-10-25
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    Ya , sure .I Will be careful during my next answer.2012-10-25
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    You can also change this one by clicking on "edit".2012-10-25
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    Correction made ... Thank You very much . This gave an opportunity to know about MAth Jax .2012-10-25