Prove or disprove: There exists an integer a such that 14 | (3a — 5) and 21 | (2a + 3).
Prove or disprove: There exists an integer a such that 14 | (3a — 5) and 21 | (2a + 3).
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

Transcribed Image Text:**Problem Statement:**
*Prove or disprove: There exists an integer \( a \) such that \( 14 \mid (3a - 5) \) and \( 21 \mid (2a + 3) \).*
**Explanation:**
The problem is asking whether there is an integer \( a \) such that the expression \( 3a - 5 \) is divisible by 14, and \( 2a + 3 \) is divisible by 21. This involves solving a system of congruences:
1. \( 3a - 5 \equiv 0 \pmod{14} \)
2. \( 2a + 3 \equiv 0 \pmod{21} \)
To solve or disprove, one could systematically approach these using techniques such as the Chinese Remainder Theorem or by finding particular solutions and verifying consistency for simultaneous congruences.
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