Seven Eleven

1. Note that $\quad11^3 \equiv 1 \pmod{7}$

2. $11^{11} \equiv (-1)^{11} \equiv 2 \pmod{3} \Longrightarrow \boxed {11^{11}=3n+2 }$

3. $11^{{11}^{11}} \equiv 11^{3n+2} \equiv (11^3)^n \cdot 11^2 \equiv 1^n \cdot 2 \equiv 2 \pmod{7}$

${11}^{{11}^{11}}$

applying fermat's little theorem,

${a}^{p-1} \equiv 1 \pmod{p}$

${a}^{6} \equiv 1 \pmod{7}$

Hence, a = 11 will work too.

${11}^{6} \equiv 1 \pmod{7}$

So, we may find ${11}^{11} mod 6$

${11}^{11} \equiv 5 \pmod{6}$

${11}^{6q + 5} = ({11}^{6q})({11}^{5})$

${11}^{5} \equiv 2 \pmod{7}$ [As ${11}^{6q} \equiv 1 \pmod{7}$ ]

The answer is 2 N=(11^11)^11 = [(7+4)^11]^11 = 7A + 4^11^11 = 7A + 4^11^11

the remainder when N is divided by 7 is the same remainder when 4^11^11 is divided by 7.

we have: 4^0=1(mod7) 4^1=4(mod7) 4^2=2(mod7) 4^3=1(mod7) so 4^n=1(mod7) if n=3k and =4(mod7) if n=3k+1 and =2(mod7) if n=3k+2. now we have to write 11^11 as a modulo of number 3 11^0=1=1mod3 11^1=11=2mod3 11^2=121=1mod3 so 11^n=1mod3 if n is pair and =2mod3 if n is odd 11 is an odd number so 11^11=2mod3 then 4^11^11 = 2(mod7) because 11^11= 3K+2

the remainder when N is divided by 7 is 2

That made me really tired