Show that $11^{10} \equiv 1 (\mod 100)$

Solution 1:

Hint:

$$(1+10)^n\equiv1+\binom n110^1\pmod{10^2}$$ for positive integer $n$

Solution 2:

Well..., you got $11^{40} \equiv 1 \pmod{100}\Rightarrow (11^{10}-1)(11^{10}+1)(11^{20}+1)\equiv 0 \pmod{100}$

Now it is easy to see that $(11^{10}+1)$ and $(11^{20}+1)$ end with $2$. Then $(11^{10}-1)\equiv 0 \pmod{25}$

Again $11^{2} \equiv 1 \pmod{4}\Rightarrow 11^{10} \equiv 1 \pmod{4}$

Since $\gcd(25,4)=1$, we will have $11^{10} \equiv 1 \pmod{100}\space\space\space\space\space\blacksquare$

Solution 3:

$11^{10}-1$

$=(1+10)^{10}-1$

$=(1+\binom{10}{1}10+\binom{10}{2}10^2+...+10^{10})-1$

$=\binom{10}{1}10+\binom{10}{2}10^2+...+10^{10}$

is divisible by $100$.