The number of coins collected per week by two coin-collectors A and B are in the ratio 3:4. If the total number of coins collected by A in 5 weeks is a multiple of 7, and the total number of coins collected by B in 3 weeks is a multiple of 24, then the minimum possible number of coins collected by A in one week is:
Question:
The number of coins collected per week by two coin-collectors A and B are in the ratio 3:4. If the total number of coins collected by A in 5 weeks is a multiple of 7, and the total number of coins collected by B in 3 weeks is a multiple of 24, then the minimum possible number of coins collected by A in one week is:
The number of coins collected per week by two coin-collectors A and B are in the ratio 3:4. If the total number of coins collected by A in 5 weeks is a multiple of 7, and the total number of coins collected by B in 3 weeks is a multiple of 24, then the minimum possible number of coins collected by A in one week is:
Options
Answer: 42
Explanation:
Let A collect 3k coins per week and B collect 4k coins per week (k a positive integer). A in 5 weeks: 5×(3k) = 15k must be a multiple of 7. 15 ≡ 1 (mod 7), so 15k ≡ k (mod 7) ⇒ k ≡ 0 (mod 7). Thus k is a multiple of 7. B in 3 weeks: 3×(4k) = 12k must be a multiple of 24. 12k ≡ 0 (mod 24) ⇒ k ≡ 0 (mod 2). Thus k is even. Combining: k is a multiple of lcm(7,2) = 14. The smallest k = 14. Therefore A collects per week: 3k = 3×14 = 42 coins
Explanation:
Let A collect 3k coins per week and B collect 4k coins per week (k a positive integer). A in 5 weeks: 5×(3k) = 15k must be a multiple of 7. 15 ≡ 1 (mod 7), so 15k ≡ k (mod 7) ⇒ k ≡ 0 (mod 7). Thus k is a multiple of 7. B in 3 weeks: 3×(4k) = 12k must be a multiple of 24. 12k ≡ 0 (mod 24) ⇒ k ≡ 0 (mod 2). Thus k is even. Combining: k is a multiple of lcm(7,2) = 14. The smallest k = 14. Therefore A collects per week: 3k = 3×14 = 42 coins
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