2^(2^(32)) mod 11 For 1420172017 mod 60 I used a logarithm rule to solve it: 1420172017 mod 60 => lg(14) a = 20172017
In essence, I calculated 20172017 mod 60 = 37 (using Euler's Theorem). Then I used that result as a power of 14: 1437 mod 60 = 44.
I figured that the same principle would also apply to 2232.
So, first I tried to get the result of 232 mod 11. Since the point of this is not to use a calculator, I simplified the term using exponentation laws: 232 = 22 mod 11 * 230 11 = 4 * 1 = 4.
I checked the result and 232 mod 11 really is 4.
The problem appears during the next step:
24 mod 11 = 5. But 2232 mod 11 = 9. Therefore, my result is wrong.
 A: Note that $2^{10}\equiv 1\pmod{11}$ (Fermat), hence if $2^{32}\equiv n\pmod{10}$, then $2^{2^{32}}\equiv 2^n\pmod{11}$.
So what is $2^{32}\bmod{10}$? 
Start by computing $2^{32}\bmod 5$. As $2^4\equiv 1\pmod 5$, clearly $2^{32}=(2^4)^8\equiv1\pmod 5$ as well. Of course $2^{32}$ is even, so (ultimately by the Chinese Remainder Theorem) $2^{32}\equiv6\pmod{10}$.
Thus we need only compute $2^6\equiv 64\equiv 9\pmod{11}$.
A: $\!\bmod 11\!:\,\ \underbrace{2^{\large 2^{\Large 32}}\!\!\equiv 2^{\large\color{#c00} 6}}_{\Large 2^{\Large\color{#0a0}{10}}\ \equiv\ 1}\,\ $ by $\ \overbrace{2^{\large 32}\!\bmod\color{#0a0}{10} = 2\underbrace{(2^{\large 31}\!\bmod 5)}_{\large 2^{\Large 4}\ \equiv\ 1}}^{\large ab\,\bmod\, ac\ \  =\ \ a(b\bmod c)\ \ \ }= 2(2^{\large 3}\!\bmod 5) = \color{#c00}6 $
A: The multiplicative order of $2$ mod $11$ is $10$, not $11$.  So you want to compute $2^{32} \bmod 10$.
A: $2^5\equiv-1\pmod{11}$
Now $2^{31}=2\cdot4^{15}=2(5-1)^{15}\equiv\equiv2(-1)\equiv3\pmod5,$
$2^{31}=5a+3$(say for some integer $a>0$)
$2^{2^{32}}=(2^{2^{31}})^2=(2^{5a+3})^2=(2^5)^{2a}\cdot2^6\equiv(-1)^{2a}\cdot2^4\equiv2^6\pmod{11}\equiv?$
