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Let $A = 2^{7^8} \hspace{2cm} B \; = \; 5^{3^{3^5}}$ $A$ is small enough that its remainder modulo $10^{10}$ can be calculated by computer: $A \equiv 7074634752 \pmod{10^{10}}$ On the other hand, $B$ is too big. We note that $B \equiv 0 \pmod{5^{10}}$ . Moreover $5^2 \equiv 1 \pmod{8}$ , so that $5^4 \equiv 1 \pmod{16}$ , $5^5 \equiv 1 \pmod{32}$ and so on, until $5^{512} \equiv 1 \pmod{2^{10}}$ . Since it is easy to calculate that $3^{3^5} \equiv 91 \pmod{512}$ , we deduce that $A \equiv 5^{91} \equiv 669 \pmod{2^{10}}$ . Using the Chinese Remainder Theorem we can deduce that $B \equiv 4423828125 \pmod{10^{10}}$ .

Thus we deduce that $A + B \equiv 7074634752 + 4423828125 \equiv \boxed{1498462877} \pmod{10^{10}}$