Question

A supplier receives orders from 5 different buyers. Each buyer places their order only on a Monday. The first buyer places the order after every 2 weeks, the second buyer, after every 6 weeks, the third buyer, after every 8 weeks, the fourth buyer, every 4 weeks, and the fifth buyer, after every 3 weeks. It is known that on January 1st, which was a Monday, each of these five buyers placed an order with the supplier.
On how many occasions, in the same year, will these buyers place their orders together excluding the order placed on January 1st?

• A. 1
• B. 5
• C. 4
• D. 2
• E. 3

Answer 1

The Correct Option is D

To solve this problem, we need to find the Least Common Multiple (LCM) of the buyers' order intervals. This will tell us after how many weeks all buyers will place orders together, excluding January 1st.
The order intervals are:

• Buyer 1: 2 weeks
• Buyer 2: 6 weeks
• Buyer 3: 8 weeks
• Buyer 4: 4 weeks
• Buyer 5: 3 weeks

Now, we find the LCM of these numbers:

1. Prime factorization:
   • 2 = 2
   • 6 = 2 × 3
   • 8 = 2³
   • 4 = 2²
   • 3 = 3
2. Take the highest power of each prime:
   • For 2: 2³
   • For 3: 3¹
3. LCM = 2³ × 3 = 8 × 3 = 24 weeks

Thus, every 24 weeks, all buyers will place orders together. Next, calculate how many 24-week intervals fit in a year excluding January 1st:

1. There are 52 weeks in a year.
2. Ordering starts on January 1st, a Monday.
3. Calculate additional occasions: 52 ÷ 24 ≈ 2 (integer division).

This means, excluding January 1st, the orders occur together twice more in the same year. Therefore, the correct answer is 2.