What's the remainder?

Fermat's little theorem states that if ${p}$ is prime then: $a^{p-1}$ $\equiv1\mod(p)$ . This can applied directly to give a remainder of $\boxed{1}$

$11^{12}=121^6$

$121 \equiv 4\mod 13$ so $121^6$ remainder is equal to $4^6$

$4^6=16^3$ as $16\equiv 3\mod 13$ , $16^3$ remainder is equal to $3^3$ remainder

Finally $\frac{27}{13}=(13)(2)+1$

Remainder $\boxed{1}$

Since $13$ is a prime number, then $\varphi(13)=13-1=12$ . Thus, $11^{12}\mod 13 = 11^{12-12}\mod 13 = \boxed{1}\mod 13$

11^{ { 12 } }=(11^{ { 2 } })^{ { 6 } }\ (11^{ { 2 } })MOD13=4\ 4^{ { 6 } }MOD13=1

11 congruent to -2(mod 13) 11^2 congruent to 4(mod 13) 11^6 congruent to 64(mod 13) 11^6 congruent to -1(mod 13) Therefore 11^12 congruent to 1(mod 13) Hence remainder: 1

11^12 mod 13 = ((13-2)^6)^2 mod 13 since (13-2)^6 mod 13 = 2^6 mod 3 = -1 mod 13. so, ((13-2)^6)^2 mod 13 = (-1)^2 mod 13 = 1 mod 13