How to find the remainders when ## 2^{50} ## and ## 41^{65} ## are?

[Math100]

Homework Statement

Find the remainders when $2^{50}$ and $41^{65}$ are divided by $7$.

Relevant Equations

None.

Consider $2^{3}=8\equiv 1 \pmod 7$.
Then $2^{50}=2^{48}\cdot 2^{2}=(2^{3})^{16}\cdot 2^{2}\equiv 1^{16}\cdot 2^{2} \pmod 7\equiv 2^{2} \pmod 7\equiv 4 \pmod 7$.
Thus $2^{50}\equiv 4 \pmod 7$.
Now observe that $41\equiv 6 \pmod 7\equiv (-1) \pmod 7$.
Then $41^{65}\equiv (-1)^{65} \pmod 7\equiv (-1) \pmod 7\equiv 6 \pmod 7$.
Thus $41^{65}\equiv 6 \pmod 7$.
Therefore, the remainders when $2^{50}$ and $41^{65}$ are divided by $7$ are $4$ and $6$.

 

[fresh_42]

Homework Statement:: Find the remainders when $2^{50}$ and $41^{65}$ are divided by $7$.
Relevant Equations:: None.

Consider $2^{3}=8\equiv 1 \pmod 7$.
Then $2^{50}=2^{48}\cdot 2^{2}=(2^{3})^{16}\cdot 2^{2}\equiv 1^{16}\cdot 2^{2} \pmod 7\equiv 2^{2} \pmod 7\equiv 4 \pmod 7$.
Thus $2^{50}\equiv 4 \pmod 7$.
Now observe that $41\equiv 6 \pmod 7\equiv (-1) \pmod 7$.
Then $41^{65}\equiv (-1)^{65} \pmod 7\equiv (-1) \pmod 7\equiv 6 \pmod 7$.
Thus $41^{65}\equiv 6 \pmod 7$.
Therefore, the remainders when $2^{50}$ and $41^{65}$ are divided by $7$ are $4$ and $6$.

Perfect.