Fun With Binomial Coefficients 2

Let $\displaystyle Z = \left ( \sum_{k=1}^5 {20 \choose 2k-1} \right )^6$ , then cosider modulo $11$

Use a few properties of Pascal Triangle, we get

$\begin{aligned} Z & \equiv & \left ( {20 \choose 1} + {20 \choose 3 } + {20 \choose 5 } + {20 \choose 7 } + {20 \choose 9 } \right )^6 \\ & \equiv & \left ( \frac {1}{2} \cdot \left ( {20 \choose 1} + {20 \choose 3 } + {20 \choose 5 } + {20 \choose 7 } + {20 \choose 9 } + {20 \choose 11} + {20 \choose 13 } + {20 \choose 15 } + {20 \choose 17 } + {20 \choose 19 } \right ) \right )^6 \\ & \equiv & 2^{(20-2) \times 6}, \text{Use Fermat's Little Theorem} \\ & \equiv & 2^{8} \\ & \equiv & 3 \\ \end{aligned}$

So $\gamma = 3$ which yields the desired answer of $\boxed{10}$