Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

My approach :

$E(7) = 6 $

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

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

share|improve this question
1  
$444^{444^{444}}(mod 7) \neq 444^{444^{444}(mod 7)}(mod 7)$ –  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
add comment

3 Answers

up vote 3 down vote accepted

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, $$444^{444^{444}} = 444^{6M} = \left( 444^6\right)^M \equiv 1^M \pmod{7} \equiv 1 \pmod{7}$$

share|improve this answer
    
Yes I meant the same.Thanks sir @marvis. –  Bazinga Jun 27 '12 at 7:26
add comment

$a^{\phi(p)}=1(\mod p)$.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 $

share|improve this answer
add comment

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 $

share|improve this answer
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.