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

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 $(1),(2)$
