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
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**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.
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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