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.
 A: Let $\displaystyle S =\binom{20}{1}+\binom{20}{3}+\binom{20}{5}+\binom{20}{7} +\binom{20}{9}\cdots (1)$
$\displaystyle S =\binom{20}{9}+\binom{20}{7}+\binom{20}{5}+\binom{20}{3}+\binom{20}{1}$
Using $\displaystyle \binom{n}{r}=\binom{n}{n-r}$
$\displaystyle S =\binom{20}{11}+\binom{20}{13}+\binom{20}{15}+\binom{20}{17}+\binom{20}{19}\cdots (2)$
Add $(1)$ and $(2)$
$\displaystyle 2S =\sum^{10}_{k=1}\binom{20}{2k-1}=2^{19}$
$$\displaystyle S =2^{18}$$
We have to calculate remainder when $\bigg(2^{18}\bigg)^6=2^{108}$ is divided by $11$
Now $2^{3}\cdot \bigg(2^{5}\bigg)^{21}=8\cdot (32)^{21}$
$=-8\cdot (1-33)^{21}=-8+M(11)=3-11+M(11)=3+M'(11)$
So remainder is $3$ when $2^{108}$ is divided by $11$
Where $M(11)$ and $M'(11)$ is Multiple of $11$
A: $$\left(\sum_{k=1}^5\binom{20}{2k-1}\right)^6=262144^6\equiv3^6\equiv3.$$
A: $$\left[\binom{20}{1}+\binom{20}{3}+\binom{20}{5}+\binom{20}{7}+\binom{20}{9}\right]^6=(20+1140+15504+77520+167960)^6\equiv 
(-2+7+5+3+1)^6\equiv3^6\equiv3\mod{11}$$
A: If it helps, the expression inside the brackets is a quarter of $2^{20}$. 
It is a half of the sum for all odd $q$; and the sum for all odd $q$ equals the sum for all even $q$.
To prove that, expand $(1-1)^{20}$ using the binomial expansion.
