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.

How can I prove using congruence that $111^{333}+333^{111}$ is divided by 7?

I tried use each factor separately but I didn't really get anywhere. Will appreciate your help.

share|improve this question
add comment

2 Answers

up vote 8 down vote accepted

$$111 \equiv -1 \pmod{7} \implies 111^{333} \equiv -1 \pmod{7}$$ $$333 \equiv 4 \pmod{7} \implies 333^{3} \equiv 1 \pmod{7} \implies 333^{111} \equiv 1 \pmod{7}$$ Hence, you get $$111^{333} + 333^{111} \equiv 0 \pmod 7$$

share|improve this answer
    
how did you get in the second line from 4 mod 7 to 1 mod 7? This is where I was stuck when trying to solve.. –  Mary Nov 29 '12 at 17:35
2  
@Nusha Marvis used $4^3=64=63+1$ –  Mark Bennet Nov 29 '12 at 18:03
add comment

$$111\equiv -1\pmod 7\implies 111^{333}\equiv (-1)^{333}=-1$$

$$333\equiv 4\pmod 7=2^2$$

Using Fermat's Little Theorem, $2^{7-1}\equiv 1\pmod 7$

$$333^{111}\equiv (2^2)^{111}\pmod 7\equiv (2^6)^{37}\equiv 1\pmod 7$$


Alternatively, $333^3\equiv 2^6\equiv1$ and $111^9\equiv-1\pmod 7$

So, $7\mid(333^3+111^9)$

But $(333^3+111^9)\mid \{(333^3)^{37}+(111^9)^{37}\}$

share|improve this answer
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.