<s>[INST] 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
 [/INST] 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 $$1^{10}+3^{10}.$$
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