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(Bhaskara, sixth century) A basket contains n eggs. If the eggs are removed 2, 3, 4, 5, or 6 at a time, then the number of eggs that remain in the basket is 1, 2, 3, 4, or 5, respectively. If the eggs are removed 7 at a time, then no eggs remain. What is the smallest number n of eggs that could have been in the basket at the start of this procedure? Hint: The first condition implies that n = 1 (mod 2).

(Bhaskara, sixth century) A basket contains n eggs. If the eggs are removed 2, 3, 4, 5, or 6 at a time, then the number of eggs that remain in the basket is 1, 2, 3, 4, or 5, respectively. If the eggs are removed 7 at a time, then no eggs remain. What is the smallest number n of eggs that could have been in the basket at the start of this procedure? Hint: The first condition implies that n = 1 (mod 2).

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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**Transcription:** 8. (Bhaskara, sixth century) A basket contains \( n \) eggs. If the eggs are removed 2, 3, 4, 5, or 6 at a time, then the number of eggs that remain in the basket is 1, 2, 3, 4, or 5, respectively. If the eggs are removed 7 at a time, then no eggs remain. What is the smallest number \( n \) of eggs that could have been in the basket at the start of this procedure? *Hint*: The first condition implies that \( n \equiv 1 \pmod{2} \). **Explanation:** This problem involves modular arithmetic, where you need to determine the smallest positive integer \( n \) that satisfies a set of congruences. The hint gives a starting point, implying that \( n \) must be odd (since \( n \equiv 1 \pmod{2} \)). To solve this problem, you would need to find \( n \) such that: - \( n \equiv 1 \pmod{2} \) - \( n \equiv 2 \pmod{3} \) - \( n \equiv 3 \pmod{4} \) - \( n \equiv 4 \pmod{5} \) - \( n \equiv 5 \pmod{6} \) - \( n \equiv 0 \pmod{7} \)
Transcribed Image Text:**Transcription:** 8. (Bhaskara, sixth century) A basket contains \( n \) eggs. If the eggs are removed 2, 3, 4, 5, or 6 at a time, then the number of eggs that remain in the basket is 1, 2, 3, 4, or 5, respectively. If the eggs are removed 7 at a time, then no eggs remain. What is the smallest number \( n \) of eggs that could have been in the basket at the start of this procedure? *Hint*: The first condition implies that \( n \equiv 1 \pmod{2} \). **Explanation:** This problem involves modular arithmetic, where you need to determine the smallest positive integer \( n \) that satisfies a set of congruences. The hint gives a starting point, implying that \( n \) must be odd (since \( n \equiv 1 \pmod{2} \)). To solve this problem, you would need to find \( n \) such that: - \( n \equiv 1 \pmod{2} \) - \( n \equiv 2 \pmod{3} \) - \( n \equiv 3 \pmod{4} \) - \( n \equiv 4 \pmod{5} \) - \( n \equiv 5 \pmod{6} \) - \( n \equiv 0 \pmod{7} \)
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