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Prove that (a) 340 - 4, and (b) 341 – 5 are divisible by 7. (You cannot rely on a calculator in your solution.) (a) 3^{40} - 4 is divisible by 7: 3^{40} = (3^{2})^{20} = 9^{20} = mod 7 = 2^{20} = (2^{2})^{10} = 4^{10} = (4^{2})^{5} = (16)^{5} = mod 7 = 2^{5} = 32 = mode 7 = 4 3^{40} -4 4-4=0 = 3^{40} 4 is divisible by 70 (b) 3^{41} - 5 is divisible by 7:

Prove that (a) 340 - 4, and (b) 341 – 5 are divisible by 7. (You cannot rely on a calculator in your solution.) (a) 3^{40} - 4 is divisible by 7: 3^{40} = (3^{2})^{20} = 9^{20} = mod 7 = 2^{20} = (2^{2})^{10} = 4^{10} = (4^{2})^{5} = (16)^{5} = mod 7 = 2^{5} = 32 = mode 7 = 4 3^{40} -4 4-4=0 = 3^{40} 4 is divisible by 70 (b) 3^{41} - 5 is divisible by 7:

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Can you help me solve only (b) 3^41 - 5 is divisible by 7
Prove that (a) 340 - 4, and (b) 341 - 5 are divisible by 7. (You cannot rely on a calculator in your solution.) (a) 3^{40} - 4 is divisible by 7: 3^{40} = (3^{2})^{20} = 9^{20} = mod 7 = 2^{20} = (2^{2})^{10} = 4^{10} = (4^{2})^{5} = (16)^{5} = mod 7 = 2^{5} = 32 = mode 7 = 4 3^{40} - 4 = 4-4=0 3^{40} - 4 is divisible by 70 (b) 3^{41} - 5 is divisible by 7:
Transcribed Image Text:Prove that (a) 340 - 4, and (b) 341 - 5 are divisible by 7. (You cannot rely on a calculator in your solution.) (a) 3^{40} - 4 is divisible by 7: 3^{40} = (3^{2})^{20} = 9^{20} = mod 7 = 2^{20} = (2^{2})^{10} = 4^{10} = (4^{2})^{5} = (16)^{5} = mod 7 = 2^{5} = 32 = mode 7 = 4 3^{40} - 4 = 4-4=0 3^{40} - 4 is divisible by 70 (b) 3^{41} - 5 is divisible by 7:
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