Remainder Problem 4

According to Fermat's little theorem $3^{16} - 1 = 17 k, 3^{181}=3^{16i}*3^5=(17k+1)^i*(17y+5)$ $\Rightarrow 3^{181} \mod{7} = \boxed{5}$

I'm gonna provide you the Basic Method :

$=3^{181}=(3^{4})^{45}\cdot 3=81^{45}\cdot 3$

Now , $81 \equiv -4 \mod{17}$

$=81^{45}\cdot 3=(-4)^{45}\cdot 3$

$=((-4)^{2})^{22}\cdot -4\cdot 3$

$=(16)^{22}\cdot -4\cdot 3$

Now , $16 \equiv -1 \mod{17}$

$=(-1)^{22}\cdot -4\cdot3=-4\cdot 3$

$=-12$

Remainder = $17-12=\boxed{5}$

From Fermat's Little Theorem, $\begin{array}{llr} &3^{181}&\mod17 \\=&3^{181\mod16}&\mod17 \\=&3^{(181-160)\mod16}&\mod17 \\=&3^{21\mod16}&\mod17 \\=&3^{21-16}&\mod17 \\=&3^5&\mod17 \\=&243&\mod17 \\=&(243-170)&\mod17 \\=&73&\mod17 \\=&(73-17)&\mod17 \\=&56&\mod17 \\=&(56-51)&\mod17 \\=&5 \end{array}$ This may be a bit long, but this is how I calculate it in my heart.