Find the remainder when 122605 is divided by 17.
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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Question
![**Problem Statement:**
Find the remainder when \(12^{2605}\) is divided by 17.
---
**Explanation:**
To solve problems of this type, typically involving large powers and finding remainders, techniques such as Fermat's Little Theorem or Euler's Theorem may be useful.
**Key Concepts:**
- **Fermat's Little Theorem**: If \(p\) is a prime number and \(a\) is any integer not divisible by \(p\), then \(a^{p-1} \equiv 1 \pmod{p}\).
In this specific problem, since 17 is a prime number:
- According to Fermat's Little Theorem:
\[
12^{16} \equiv 1 \pmod{17}
\]
Thus, to find \(12^{2605} \mod 17\), first determine the power modulo 16 because of Fermat's Theorem:
\[
2605 \div 16 = 162 \text{ remainder } 13
\]
This means \(2605 \equiv 13 \pmod{16}\).
Therefore:
\[
12^{2605} \equiv 12^{13} \pmod{17}
\]
Calculating progressively:
\[
12^1 \equiv 12 \pmod{17}
\]
\[
12^2 \equiv 144 \equiv 8 \pmod{17}
\]
\[
12^4 \equiv (12^2)^2 \equiv 8^2 \equiv 64 \equiv 13 \pmod{17}
\]
\[
12^8 \equiv (12^4)^2 \equiv 13^2 \equiv 169 \equiv 16 \equiv -1 \pmod{17}
\]
Thus, \(12^{13} = 12^8 \cdot 12^4 \cdot 12 \equiv (-1) \cdot 13 \cdot 12\).
Continuing:
\[
(-1) \cdot 13 \equiv -13 \equiv 4 \pmod{17}
\]
\[
4 \cdot 12 \equiv 48 \equiv 14 \pmod{17}
\]
Thus, the remainder when \(12^{2605](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2a4399a9-5724-42c8-89a0-9bc27dd1a0f2%2Fe43034d4-e774-4fab-8128-3fa0322e43fd%2Fks2pwc_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the remainder when \(12^{2605}\) is divided by 17.
---
**Explanation:**
To solve problems of this type, typically involving large powers and finding remainders, techniques such as Fermat's Little Theorem or Euler's Theorem may be useful.
**Key Concepts:**
- **Fermat's Little Theorem**: If \(p\) is a prime number and \(a\) is any integer not divisible by \(p\), then \(a^{p-1} \equiv 1 \pmod{p}\).
In this specific problem, since 17 is a prime number:
- According to Fermat's Little Theorem:
\[
12^{16} \equiv 1 \pmod{17}
\]
Thus, to find \(12^{2605} \mod 17\), first determine the power modulo 16 because of Fermat's Theorem:
\[
2605 \div 16 = 162 \text{ remainder } 13
\]
This means \(2605 \equiv 13 \pmod{16}\).
Therefore:
\[
12^{2605} \equiv 12^{13} \pmod{17}
\]
Calculating progressively:
\[
12^1 \equiv 12 \pmod{17}
\]
\[
12^2 \equiv 144 \equiv 8 \pmod{17}
\]
\[
12^4 \equiv (12^2)^2 \equiv 8^2 \equiv 64 \equiv 13 \pmod{17}
\]
\[
12^8 \equiv (12^4)^2 \equiv 13^2 \equiv 169 \equiv 16 \equiv -1 \pmod{17}
\]
Thus, \(12^{13} = 12^8 \cdot 12^4 \cdot 12 \equiv (-1) \cdot 13 \cdot 12\).
Continuing:
\[
(-1) \cdot 13 \equiv -13 \equiv 4 \pmod{17}
\]
\[
4 \cdot 12 \equiv 48 \equiv 14 \pmod{17}
\]
Thus, the remainder when \(12^{2605
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