Show that $2222^{5555} + 5555^{2222}$ is divisible by $ 7$ Show that $2222^ {5555} + 5555 ^ {2222}$ is divisible by $7$. I tried factorizing but it didnt lead to anything. Can divisibility rules be used? Any ideas please tell me.
 A: Use modular arithmetic, plus some remarks on the particular numbers involved:
First note that $5555=55\times101$ and $2222=22\times101$, hence


*

*$5555\equiv -1\times 3=-3 \mod 7,\quad 5555\equiv 1\times -1=-1 \mod 6$

*$2222\equiv 1\times 3=3 \mod 7,\quad 2222 \equiv 4\times -1 \equiv 2 \mod 6$
By Little Fermat,  $x^{5555}\equiv x^{5555\bmod 6}=x^{-1}$,   $\quad x^{2222}\equiv x^{2222\bmod 6}=x^{2}$. Thus
$$ 2222^{5555}+5555^{2222}\equiv 3^{-1}+(-3)^2\equiv 5 +2 \equiv 0 \mod 7.$$
A: You may use modular arithmetic for such questions involving huge numbers.
$$2222^{5555} ≡ 3^{2222} \pmod{7}.$$
Since $$3^6 ≡ 729 ≡ 1\pmod{7}, 3^{2222} ≡ 3^{6*925} * 3^5 ≡ 243 ≡ 5 \pmod{7}.$$
Also 
$$5555^{2222} ≡ 4^{2222} \pmod{7}$$
$$4^3 ≡ 64 ≡ 1 \pmod{7}, 4^{2222} ≡ 4^{3*740}*4^2 ≡ 16 ≡ 2\pmod{7}.$$
Adding them up, $$5555^{2222} + 2222^{5555} ≡ 2+5 ≡ 7 ≡ 0 \pmod{7},$$
hence proving the divisibility.
A: Compute the remainder that arise when 2222 is divided by 7; also you can compute the remainder for 5555. Now, let be $2222=7p_1+q_1, 5555=7p_2+q_2$ then it follows:
$2222^{5555}+5555^{2222}=(7p_1+q_1)^{5555}+(7p_2+q_2)^{2222}$.
When you expand out the binomial you will see that the only term that could not be divisible by 7 is $q_1^{5555}+q_2^{2222}$. Here, $q_1,q_2$ are your remainders; now you can prove that $q_1^{5555}+q_2^{2222}$ is divisible by 7.
A: You could use Fermat's little theorem, which in here basically says that, for all integers $a$ and $n$, $a^n,a^{n+6},a^{n+12},a^{n+18},\dotsc$ will all have the same remainder when divided by 7.
\begin{aligned}
(2222^{5555} + 5555^{2222}) \bmod 7 
&= \bigl((2222 \bmod 7)^{5555 \bmod 6} + (5555 \bmod 7)^{2222 \bmod 6}\bigr) \bmod 7 \\
&= (3^5 + 4^2) \bmod 7 \\
&= 259 \bmod 7 \\
&= 0
\end{aligned}
A: ${\rm mod}\ 7\!:\,\ \overbrace{\color{#90f}4^{2222}\!+(\color{#0a0}{-4})^{5555}}^{\large \color{#90f}{5555}^{2222}+\:\!\color{#0a0}{2222}^{5555}} \equiv 4^{2222}(\color{#c00}1-\overbrace{4^{3333}}^{\large (\color{#c00}{2^6})^{1111}\!\!\!})\equiv 0\,$ by $\,\color{#c00}{1\equiv 2^6}$
