Show that the integer $111^{333} + 333^{111}$ is divisible by $7$.

 Example: Show that the integer $111^{333} + 333^{111}$ is divisible by $7$.

Answer: Firstly, we will find the remainder when $111^{333} $ and $ 333^{111}$ is divided by $7$. Then with help of property $4.$ of Theorem, we show that  the integer $111^{333} + 333^{111}$ is divisible by $39$.

We have $111 \equiv 6 ~(mod ~7)$ but $6 \equiv -1 ~(mod ~7) $. Therefore $111 \equiv -1 ~(mod ~7) $. By property $6.$ of Theorem, we have  $111^{2} \equiv 1 ~(mod ~7)$. We can write $333$ as $333 = 166 \cdot 2 + 1$. Therefore $111^{333} \equiv (111^{2})^{166} \cdot 111 \equiv 1^{166} \cdot 111 \equiv  6 ~(mod ~7)$, i.e. $111^{333} \equiv  6 ~(mod ~7)$. This gives that $6$ is a remainder when $111^{333}$ is divided by $7$.

We have $333 \equiv 4 ~(mod ~7)$.  By property $6.$ of Theorem, we have $333^2 \equiv 4^2 \equiv 2 ~(mod ~7)$. Again, with the help of property $6.$ of Theorem, we have  $(333^2)^3 = 333^6 \equiv 2^3 \equiv 1 ~(mod ~7)$. We can write $111$ as $111 = 18 \cdot 6 + 3$. Therefore $333^{111} \equiv (333^{6})^{18} \cdot 333^3 \equiv 1^{18} \cdot 333^3 \equiv  333^3 \equiv 4^3 \equiv 1 ~(mod ~7)$, i.e. $333^{111} \equiv  1 ~(mod ~7)$. This gives that $1$ is a remainder when $333^{111}$ is divided by $7$.

By property $4.$ of Theorem, we have $111^{333} + 333^{111} \equiv 6 + 1 \equiv 0 ~(mod ~ 7)$. This show that  $0$ is a remainder when the integer $111^{333} + 333^{111}$ is divided by $7$, i.e. the integer $111^{333} + 333^{111}$ is divisible by $7$.

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