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# Modulo Arithmetic

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Hi, could someone help me with this problem?

Given $$a \equiv 1 \pmod{7}$$$$b \equiv 2 \pmod{7}$$, and $$c \equiv 6 \pmod{7}$$, what is the remainder when $$a^{81} b^{91} c^{27}$$ is divided by 7?

Thank you so much!

Oct 25, 2020

#1
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By inspection, the simplest solution is:

a =1,  b = 2  and  c = 6 and: [1^81*2^91*6^27] mod 7 = 5

The full solution is: a = 7 + 1 =8,  b =7 + 2 =9  and  c =7  + 6 =13  and [8^81*9^91*13^27] mod 7 = 5

Oct 25, 2020
#2
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Thank you so much!

Caffeine  Oct 28, 2020