Inspired by Ayush Rai

Note that, $95^2=25 \mod(100)$ and $25^2= 25 \mod (100)$ . Thus, $95^{1999} = 95 \times \big( 95^2 \times 95^2 \times \ldots \times 95^2\big) \mod(100)= 95\times 25 = 75 \mod(100)$ Again, $96^5 = 76 \mod(100)$ and $76^2= 76 \mod (100)$ . Thus, $96^{1999}= 96^4 \times \big( 96^5 \times 96^5 \times \ldots \times 96^5 \big) \mod (100)= 56\times 76 \mod(100)= 56 \mod(100)$ Thus the answer is $75+56 \mod(100)= 31$ . $\blacksquare$

1.- $95^{1999} \equiv (-5)^{1999} \pmod{100}$ Now, it can be proved by induction that $\forall n\ge 2, \space 5^n$ ends in $25$ . So

$95^{1999} \equiv (-5)^{1999} \equiv -25 \equiv 75 \pmod{100}$

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2.- $96 \equiv 0 \pmod{4} \Rightarrow 96^{1999} \equiv 0 \pmod 4$ . I'm going now to use Euler theorem . ( $\phi (25) = 20$ ) and due to $21$ and $25$ are coprime $21^{20} \equiv 1 \pmod{25} \Rightarrow$ $96^{1999} \equiv 21^{1999} = (21^{20})^{99} \cdot 21^{19} \equiv 21^{19} \equiv - 4^{19} = - (4^5)^3\cdot 4^4 \equiv 6 \pmod{25}$ ( $4^5 \equiv -1 \pmod{25}$ ). In this point, we can use Chinese Remainder theorem $\Rightarrow 96^{1999} \equiv 4\cdot 19 \cdot 6 + 0 \cdot 25 \cdot 1 = 456 \equiv 56 \pmod{100}$ . Therefore, $95^{1999} + 96^{1999} \equiv 75 + 56 \equiv 31 \pmod{100}$