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A man offers 2 flowers on even days, 3 flowers on odd days divisible by 3 and offers 2 extra flowers on the days divisible by 5. How many flowers did he offer in April 1984?

Question

A man offers 2 flowers on even days, 3 flowers on odd days divisible by 3 and offers 2 extra flowers on the days divisible by 5. How many flowers did he offer in April 1984?

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Solution

To solve this problem, we need to consider the number of even days, odd days divisible by 3, and days divisible by 5 in April 1984.

  1. April has 30 days. Half of these days are even, so there are 15 even days. On even days, the man offers 2 flowers. So, on even days, he offers 15 * 2 = 30 flowers.

  2. April has 15 odd days. Among these, the days divisible by 3 are 3, 9, 15, 21, and 27. So, there are 5 odd days divisible by 3. On these days, the man offers 3 flowers. So, on odd days divisible by 3, he offers 5 * 3 = 15 flowers.

  3. The days in April that are divisible by 5 are 5, 10, 15, 20, 25, and 30. So, there are 6 days divisible by 5. On these days, he offers 2 extra flowers. So, on days divisible by 5, he offers 6 * 2 = 12 extra flowers.

Adding these up, the total number of flowers the man offers in April 1984 is 30 (from even days) + 15 (from odd days divisible by 3) + 12 (extra flowers on days divisible by 5) = 57 flowers.

This problem has been solved

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