78.1k views
3 votes
Problem 7, please answer the following step by step.

Problem 7, please answer the following step by step.-example-1
User GarryOne
by
8.2k points

1 Answer

2 votes

(a) This is the same as computing 4⁵⁵ (mod 11). We have

4² ≡ 16 ≡ 5 (mod 11)

4³ ≡ 4 • 5 ≡ 20 ≡ 9 (mod 11)

4⁴ ≡ 4 • 9 ≡ 36 ≡ 3 (mod 11)

4⁵ ≡ 4 • 3 ≡ 12 ≡ 1 (mod 11)

Then from here,

4⁵⁵ ≡ (4⁵)¹⁰ • 4⁵ ≡ 1¹⁰ • 1⁵ ≡ 1 (mod 11)

(b) Each term in the sum

4ⁿ + 4ⁿ⁺¹ + 4ⁿ⁺² + 4ⁿ⁺³ + 4ⁿ⁺⁴

has a common factor of 4ⁿ, so this sum is the same as

4ⁿ (1 + 4 + 16 + 64 + 256) = 4ⁿ • 341 = 4ⁿ • 11 • 31

So the sum is indeed divisible by 11 for all integers n.

(c) Since 4ⁿ = (2ⁿ)², we know the sum is also divisible by 11 when a = 2.

User Steve Hill
by
7.2k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories