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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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+1

Thank you so much!

Caffeine  Oct 28, 2020

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