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:

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
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Chapter10: Sequences, Series, And Probability
Section10.5: The Binomial Theorem
Problem 16E
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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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