Determining the last two digits of $229^{10} +37^{10}$ Determine the last two digits of: $229^{10}+37^{10}.$
I do not want to use the Euler-totient function or the carmichael function please!
Thanks
 A: Hint:
you can see that $229^{10}=(220+9)^{10}\equiv 9^{10}\equiv729^3\cdot9\equiv29^3\cdot9\equiv 1\pmod{100}$,
also $37^{10}\equiv 7^{10}\equiv49\pmod{100}$, 
hence $229^{10}+37^{10}\equiv 50\pmod{100}.$
A: Applying binomial theorem, one can rewrite
$$
229^{10}+37^{10} = (230-1)^{10} + (40-3)^{10} \\
=230^{10} - \binom{10}{1}230^9+\binom{10}{2}230^8- \ldots +\binom{10}{8}230^2-\binom{10}{9}230+1\\
+40^{10} - \binom{10}{1}40^9 3+\binom{10}{2}40^8 3^2- \ldots +\binom{10}{8}40^2 3^8-\binom{10}{9}40\cdot 3^9+3^{10}\\
=(......00 -10\cdot230+1)+(.......00-10\cdot40\cdot3^9+3^{10}),
$$
so the last two digits of $$229^{10}+37^{10}$$ are the same as the last two digits of $$1^{10}+3^{10}.$$
A: $37^{10}\equiv 7^{10}\equiv 49\pmod{100}$
$229^{10}\equiv 9^{10}\pmod {100}$
$\begin{align}9^3\equiv 29\pmod {100}, 9^2\equiv-19\pmod{100} & \implies 9^5\equiv -(29\cdot19)\pmod{100}\\&\implies 9^5\equiv -551\pmod{100}\\&\implies 9^5\equiv -51\pmod{100}\\&\implies 9^{10}\equiv 1\pmod{100}\end{align}$
A: We need $229^{10}+37^{10}\pmod{100}$
As $100=4\cdot25$ where $(4,25)=1$
$229\equiv1\pmod4\implies229^{10}\equiv1;37\equiv1\pmod4\implies37^{10}\equiv1$
$\implies229^{10}+37^{10}\equiv1+1\pmod4\equiv2\ \ \ \ (1)$
Again, $229\equiv4\pmod{25}\implies229^{10}\equiv4^{10}$
But $4^{10}=(2^2)^{10}=2^{20}$
and $37\equiv12\pmod{25}\equiv2^23\implies37^{10}\equiv(2^23)^{10}\equiv2^{20}3^{10}$
$\implies229^{10}+37^{10}\equiv2^{20}(1+3^{10})\pmod{25}$
Again, $3^3=27\equiv2\pmod{25}\implies3^{10}=3(3^3)^3\equiv3(2)^3\pmod{25}\equiv-1$
$\implies229^{10}+37^{10}\equiv1-1\pmod{25}\equiv0\ \ \ \ (2)$
Now apply Chinese Remainder Theorem on $(1),(2)$
