Show that $41$ divide $2^{20} - 1$. OR Show that $1$ is remainder when $2^{20}$ is divided by $41$. OR show that $2^{20} \equiv 1 ~(mod ~41)$.

 Example:  Show that $41$ divide $2^{20} - 1$.  

OR  

Show that $1$ is remainder when $2^{20}$ is divided by $41$. 

OR  

Show that $2^{20} \equiv 1 ~(mod ~41)$.


Answer: We have to show that $41$ divide $2^{20} - 1$. Clearly we have $2 \equiv 2 ~(mod ~41) $. By property $6.$ of Theorem, we have  $2^{2} \equiv 4 ~(mod ~41)$. On the same line we have $2^{3} \equiv 8 ~(mod ~41)$. Similarly,  $2^{4} \equiv 16 ~(mod ~41)$ and $2^{5} \equiv 32 ~(mod ~41)$. But $32 \equiv -9 ~(mod ~41)$. Therefore $2^{5} \equiv -9 ~(mod ~41)$. Hence $2^{5} \cdot 2^{5} \equiv (-9) \cdot (-9) ~(mod ~41)$, i.e. $2^{10} \equiv 81 ~(mod ~41)$. But $81 \equiv -1 ~(mod ~41) $, so $2^{10} \equiv -1 ~(mod ~41)$. Again, we have  $2^{10} \cdot 2^{10} \equiv (-1) \cdot (-1) ~(mod ~41)$, i.e. $2^{20} \equiv 1 ~(mod ~41)$. Hence $41$ divide $2^{20} - 1$.

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