Remainder

Fermat's Little Theorem: ${a}^{p}\equiv a\pmod{p}$ , $p$ prime and $a$ integer

${2}^{70}\equiv {2}^{13}*2^{13}*2^{13}*2^{13}*2^{13}*2^5\equiv 2^5*2^5\equiv 6*6\equiv 10\pmod{13}$

${3}^{70}\equiv 3^{13}*3^{13}*3^{13}*3^{13}*3^{13}*3^5\equiv 3^5*3^5\equiv 9*9\equiv 3\pmod{13}$

$2^{70}+3^{70}\equiv 10+3\equiv 0\pmod{13}$

From Fermat's little theorem

$2^{13} \equiv 2 \pmod{13}$

$\implies 2^{14} \equiv 2^2 \pmod{13}$

$\implies (2^{14})^5 \equiv (2^2)^{5} \pmod{13}$

$\implies 2^{70} \equiv 2^{10} \equiv 2^4*2^4*2^2 \equiv 3*3*4 \equiv 10 \pmod{13}$

Similarly,

$3^{70} \equiv 3^{10} \equiv 3^3*3^3*3^3*3 \equiv 1*1*1*3 \equiv 3 \pmod{13}$

Therefore,

$2^{70}+3^{70} \equiv 10 +3 \equiv 0 \pmod{13}$

a^n + b^n is divisable by a + b only when n is odd. 2^70 + 3^ 70 = 4^35 + 9^35 So 4+9=13... Therefore it completely divisable by 13. Answer is zero.

Using Fermat's Little Theorem (which states that $a^{p} \equiv {a} \mod\quad{p}$ , we have $2^{70} \equiv {{2}^{65}} * {{2}^{5}} \equiv {{2}^{5}}^{13} * {2}^{5} \equiv {{32}^{13}} * {32} \equiv {32} * {32} \equiv {{6}^{2}} \equiv {36} \equiv {-3} \quad\mod 13$

In similar manner

${3}^{70} \equiv {{3}^{5}}^{13} * {3}^{5} \equiv {243}^{2} \equiv {-4}^{2} \equiv {16} \equiv {3} \quad\mod{13}$

Adding together results in ${-3} + {3} \equiv \boxed{ 0} \quad\mod{13}$

${ 2 }^{ 70 }\equiv { 1024 }^{ 7 }\equiv { (1014+10) }^{ 7 }\equiv { 10 }^{ 7 }\equiv { (13-3) }^{ 7 }\equiv -2187\equiv 10(mod\quad 13)\\ { 3 }^{ 70 }\equiv { 59049 }^{ 7 }\equiv { (59046+3) }^{ 7 }\equiv 2187\equiv 3(mod\quad 13)\\ { 2 }^{ 70 }+{ 3 }^{ 70 }\equiv 10+3\equiv 0(mod\quad 13)$

nice solution

2^70=2^72 * 2^-2(mod 13)=2^-2=4 3^70=3^72 * 3^-2(mod 13)=3^-2=9 then 2^70+3^70(mod 13)=4+9=13(mod 13)=1