Linear Algebra: A Modern Introduction

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First, note that: $2^{100} = (2^{10})^{10}=((2^5)^2)^{10}$ In $\mathbb{Z}_11$, $2^5 = 10$, since $32$ divided by $11$ gives a remainder of $10$. Thus, we have: $(10^2)^{10}$ Now, note that $10^2=100$ is simply $1$ in $Z_11$, since $100=9*11 + 1$. Thus, the exponent is equal to $1^{10}=1$.