The remainder when $2^{70}+3^{70}$ is divided by 13 is:

Powers of 2 repeat mod 13 with period 12, and $70=12\times 5+10$, so $2^{70}$ ≡ $2^{10}=1024$ ≡ 10 (mod 13). Powers of 3 repeat with period 3 since $3^3=27$ ≡ 1, and $70=3\times 23+1$, so $3^{70}$ ≡ 3. The sum is 10 + 3 = 13 ≡ 0, so the remainder is 0.