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Question
Question: For n>= 1(n is greater or equal to 1) use congruence theory to
establish:
27|2^5n+1 + 5^5n+2.
I tried breaking it up to 27|2^5n(2^1) + 5^5n(5^2). This brings me to
27|32^n(2) + 5^5n(25). Then I break it up more by 3*9|32^n(2) +
5^5n(25).
Now I am stuck. Please help.
Hello JJ
I am afraid there is a misprint somewhere. it doesn't work when n=1. The expression evaluates to 78189 which is congruent to 24 mod 27. I am assuming that 5n+1 and 5n+2 should be parenthesised.
When you find out the misprint, the following could be useful:
5^2 = -2 mod 27. 2^5 = 5 mod 27.
2^18 = 5^18 = 1 mod 27 from Euler's theorem (Phi(27) = 18.
Best wishes
vijilant
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Vijilant
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Most questions on number theory, divisibility, primes, Euclidean algorithm, Fermat`s theorem, Wilson`s theorem, factorisation, euclidean algorithm, diophantine equations, Chinese remainder theorem, group theory, congruences, continued fractions.
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Teacher of math for 50 years
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