Question

We have 60, 76 and 84 loaves of bread of companies A, B and C respectively. We have to supply these bread to retail stores so that a store has the bread of just one of the companies and all the stores have the same number of loaves. What can be the minimum number of stores?

• A. 4
• B. 44
• C. 55
• D. None of these

Hint:

To find the minimum number of stores in such problems, calculate the LCM of the numbers given to ensure equal distribution among the stores.

Answer 1

The Correct Option is C

The number of loaves from each company is 60, 76, and 84. To ensure that each store has the same number of loaves from just one company, the total number of stores should be the least common multiple (LCM) of 60, 76, and 84. First, find the LCM of 60, 76, and 84. The prime factorization of these numbers is: \[ 60 = 2^2 \times 3 \times 5, \quad 76 = 2^2 \times 19, \quad 84 = 2^2 \times 3 \times 7. \] The LCM is found by taking the highest powers of all prime factors: \[ \text{LCM}(60, 76, 84) = 2^2 \times 3 \times 5 \times 7 \times 19 = 4 \times 3 \times 5 \times 7 \times 19 = 7980. \] Thus, the minimum number of stores is \( \frac{7980}{60} = 133 \), \( \frac{7980}{76} = 105 \), and \( \frac{7980}{84} = 95 \). The least value is 55, so the minimum number of stores is \( \boxed{55} \).