Numbers numbers

$\begin{aligned} 11^{22} & = (10+1)^{22} \\ & = 10^{22} +...+\begin{pmatrix} 22 \\ 21 \end{pmatrix} 10 \dot{} 1^{21} + 1^{22} \\ & = 10^{22} +...+22\dot{}10\dot{}1 + 1^{22} \\ & = ...221 \end{aligned}$

The ten's digit is therefore $\boxed{2}$ .

11^2 = 121 11^3 = 12321 11^4 = 1234321 etc. Notice how every time, the tens in always 2.

11²²=121¹¹=21¹¹=21x441^5=21x41^5= 21x41*1681²=861x81²=61x6561=61²= 3721=21 (mod 100) so our digit is 2 Also 11ⁿ gives reminders 11,21,31,41,51,61,71,81,91,01 when divided by 100 and repeating every 10th,so 22=2(mod 10) so 11²²=21(mod 100) so 2 is answer