Modulus Congruence ProofDate: 04/18/2001 at 03:11:48 From: Ooi Chin Wah Subject: Number theory Thanks for your answers to the two questions that I asked earlier. I would be very grateful if you can help me again. Prove 2^(3n+2)+21n = 4 mod (49) Thanks. Date: 04/18/2001 at 10:01:54 From: Doctor Rob Subject: Re: Number theory Thanks for writing back to Ask Dr. Math. First let's verify the congruence modulo 7. We know 2^3 = 1 (mod 7). That implies that: 2^(3*n+2) = 4*(2^3)^n = 4*1^n = 4 (mod 7) 2^(3*n+2) + 21*n = 4 (mod 7) and the congruence holds modulo 7. Now we know that: 2^3 = 1 + 7 2^(3*n) = (1+7)^n Expand using the Binomial Theorem, and reduce modulo 49 = 7^2. Then multiply both sides by 4, and rearrange to get your answer. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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