Asked by Laura
find remainders when 2^50 and 41^65 are divided by 7. this is a mod problem but I don't know how to work it...
Answers
Answered by
Count Iblis
By Fermat's little theorem we have:
a^6 = 1
(equality means equality mod 7)
if GD(a,7) = 1.
2^48 is thus equal to 1, therefore
2^50 = 2^2 = 4
To compute 41^65, we can use that
41 = -1, and taking the 65th power then gives:
(-1)^65 = -1 = 6
a^6 = 1
(equality means equality mod 7)
if GD(a,7) = 1.
2^48 is thus equal to 1, therefore
2^50 = 2^2 = 4
To compute 41^65, we can use that
41 = -1, and taking the 65th power then gives:
(-1)^65 = -1 = 6
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