# Find the remainder in the following case.

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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$444^{444^{444}}(mod 7) \neq 444^{444^{444}(mod 7)}(mod 7)$ –  Aang Jun 27 '12 at 7:18
If by E(7) you mean $\varphi(7)=6$, then yeah. –  anon Jun 27 '12 at 7:18
@avatar: But OP is doing $\bmod6$ in the exponent, not $\bmod7$. See Euler's theorem. –  anon Jun 27 '12 at 7:19
Sorry!,i didn't notice that. –  Aang Jun 27 '12 at 7:31

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. –  Bazinga Jun 27 '12 at 7:26

$a^{\phi(p)}=1(\mod p)$.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, here, here and here. –  Julian Kuelshammer Oct 25 '12 at 16:52
Ya , sure .I Will be careful during my next answer. –  Harish Kayarohanam Oct 25 '12 at 18:20
You can also change this one by clicking on "edit". –  Julian Kuelshammer Oct 25 '12 at 18:22
Correction made ... Thank You very much . This gave an opportunity to know about MAth Jax . –  Harish Kayarohanam Oct 25 '12 at 19:40