The remainder when $( \sum_{k=1}^5 { ^{20} \mathrm C (2k-1) } )^6$ is divided by 11, is : How do I approach this? Any help is appreciated.

$$\left[\binom{20}{1}+\binom{20}{3}+\binom{20}{5}+\binom{20}{7}+\binom{20}{9}\right]^6=(20+1140+15504+77520+167960)^6\equiv 
(-2+7+5+3+1)^6\equiv3^6\equiv3\mod{11}$$
