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5. Find all integers that have a remainder of 1 when divided by 3, 5, and 7.

5. Find all integers that have a remainder of 1 when divided by 3, 5, and 7.

Advanced Engineering Mathematics
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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**Problem 5:** Find all integers that have a remainder of 1 when divided by 3, 5, and 7. --- **Explanation:** This problem involves finding integers that satisfy a specific condition with respect to division by the numbers 3, 5, and 7. Specifically, we are looking for integers \( x \) that meet the following set of congruences: 1. \( x \equiv 1 \pmod{3} \) 2. \( x \equiv 1 \pmod{5} \) 3. \( x \equiv 1 \pmod{7} \) This means that \( x \), when divided by each of 3, 5, and 7, leaves a remainder of 1. --- **Solution:** To solve this, observe that if \( x \equiv 1 \pmod{3} \), \( x \equiv 1 \pmod{5} \), and \( x \equiv 1 \pmod{7} \), then \( x \) must be 1 more than a multiple of the least common multiple of 3, 5, and 7. 1. Find the Least Common Multiple (LCM) of 3, 5, and 7: - \( \text{LCM}(3, 5, 7) = 3 \times 5 \times 7 = 105 \). 2. All integers of the form: - \( x = 105k + 1 \) for any integer \( k \), will satisfy the condition that they leave a remainder of 1 when divided by 3, 5, and 7. Thus, the integers that satisfy the given conditions are those that can be expressed in this form.
Transcribed Image Text:**Problem 5:** Find all integers that have a remainder of 1 when divided by 3, 5, and 7. --- **Explanation:** This problem involves finding integers that satisfy a specific condition with respect to division by the numbers 3, 5, and 7. Specifically, we are looking for integers \( x \) that meet the following set of congruences: 1. \( x \equiv 1 \pmod{3} \) 2. \( x \equiv 1 \pmod{5} \) 3. \( x \equiv 1 \pmod{7} \) This means that \( x \), when divided by each of 3, 5, and 7, leaves a remainder of 1. --- **Solution:** To solve this, observe that if \( x \equiv 1 \pmod{3} \), \( x \equiv 1 \pmod{5} \), and \( x \equiv 1 \pmod{7} \), then \( x \) must be 1 more than a multiple of the least common multiple of 3, 5, and 7. 1. Find the Least Common Multiple (LCM) of 3, 5, and 7: - \( \text{LCM}(3, 5, 7) = 3 \times 5 \times 7 = 105 \). 2. All integers of the form: - \( x = 105k + 1 \) for any integer \( k \), will satisfy the condition that they leave a remainder of 1 when divided by 3, 5, and 7. Thus, the integers that satisfy the given conditions are those that can be expressed in this form.
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