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

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To find the remainders of 2^50 and 41^65 when divided by 7, the calculations show that 2^3 is congruent to 1 modulo 7, leading to 2^50 being congruent to 4 modulo 7. For 41, since it is congruent to 6 modulo 7, raising it to the 65th power results in 41^65 being congruent to 6 modulo 7. Therefore, the remainders are 4 for 2^50 and 6 for 41^65. The final results confirm the calculations are accurate.
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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 ##.
 
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Math100 said:
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.
 
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