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Find the last digit of the hexidecimal (base 16) expansion of 91005

Find the last digit of the hexidecimal (base 16) expansion of 91005

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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**Problem Statement:** Find the last digit of the hexadecimal (base 16) expansion of \( 9^{1005} \). **Solution Approach:** To solve this problem, we need to find the remainder when \( 9^{1005} \) is divided by 16, as this will give us the last digit in the hexadecimal representation. Using Euler's theorem, which states that if \( a \) and \( n \) are coprime, then \( a^{\phi(n)} \equiv 1 \pmod{n} \), we proceed with the following steps: 1. **Find \(\phi(16)\):** - The Euler's totient function, \(\phi(16) = 16 \times (1 - \frac{1}{2}) = 8\). 2. **Apply Euler's Theorem:** - Since 9 and 16 are coprime, we have \( 9^8 \equiv 1 \pmod{16} \). 3. **Simplify \(9^{1005} \mod 16\):** - \( 1005 \div 8 = 125\) remainder \(5\). - Therefore, \( 9^{1005} \equiv 9^5 \pmod{16} \). 4. **Calculate \(9^5 \mod 16\):** - \( 9^2 = 81 \equiv 1 \pmod{16} \) (since \( 81 \div 16 = 5\) remainder \(1\)). - \( 9^4 = (9^2)^2 = 1^2 \equiv 1 \pmod{16} \). - \( 9^5 = 9^4 \times 9 = 1 \times 9 \equiv 9 \pmod{16} \). Thus, the last digit of the hexadecimal expansion of \( 9^{1005} \) is 9.
Transcribed Image Text:**Problem Statement:** Find the last digit of the hexadecimal (base 16) expansion of \( 9^{1005} \). **Solution Approach:** To solve this problem, we need to find the remainder when \( 9^{1005} \) is divided by 16, as this will give us the last digit in the hexadecimal representation. Using Euler's theorem, which states that if \( a \) and \( n \) are coprime, then \( a^{\phi(n)} \equiv 1 \pmod{n} \), we proceed with the following steps: 1. **Find \(\phi(16)\):** - The Euler's totient function, \(\phi(16) = 16 \times (1 - \frac{1}{2}) = 8\). 2. **Apply Euler's Theorem:** - Since 9 and 16 are coprime, we have \( 9^8 \equiv 1 \pmod{16} \). 3. **Simplify \(9^{1005} \mod 16\):** - \( 1005 \div 8 = 125\) remainder \(5\). - Therefore, \( 9^{1005} \equiv 9^5 \pmod{16} \). 4. **Calculate \(9^5 \mod 16\):** - \( 9^2 = 81 \equiv 1 \pmod{16} \) (since \( 81 \div 16 = 5\) remainder \(1\)). - \( 9^4 = (9^2)^2 = 1^2 \equiv 1 \pmod{16} \). - \( 9^5 = 9^4 \times 9 = 1 \times 9 \equiv 9 \pmod{16} \). Thus, the last digit of the hexadecimal expansion of \( 9^{1005} \) is 9.
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First we need to find out what is the last digit of 91005 in decimal ( base 10) system. Then we convert the last digit from decimal system to hexadecimal system.

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