There are 48 cricket balls, 72 hockey balls and 84 tennis balls and they have to be arranged in several rows in such a way every row contains the same number of balls of one type. What is the minimum number of rows required for this to happen?

Options

A. 12
B. 16
C. 17
D. 19

Answer

To solve the problem of arranging 48 cricket balls, 72 hockey balls, and 84 tennis balls into rows such that each row contains the same number of balls of one type, we need to find the highest common factor (HCF) of the three numbers. Here’s a step-by-step solution:

### Step 1: Prime Factorization
First, we will find the prime factorization of each number.

- **For 48:**
- 48 can be divided by 2:
- 48 ÷ 2 = 24
- 24 ÷ 2 = 12
- 12 ÷ 2 = 6
- 6 ÷ 2 = 3
- 3 is a prime number.
- Therefore, the prime factorization of 48 is \(2^4 \times 3^1\).

- **For 72:**
- 72 can be divided by 2:
- 72 ÷ 2 = 36
- 36 ÷ 2 = 18
- 18 ÷ 2 = 9
- 9 ÷ 3 = 3
- 3 is a prime number.
- Therefore, the prime factorization of 72 is \(2^3 \times 3^2\).

- **For 84:**
- 84 can be divided by 2:
- 84 ÷ 2 = 42
- 42 ÷ 2 = 21
- 21 ÷ 3 = 7
- 7 is a prime number.
- Therefore, the prime factorization of 84 is \(2^2 \times 3^1 \times 7^1\).

### Step 2: Identify Common Factors
Next, we will identify the common prime factors from the factorizations:

- **Common prime factors:**
- The common prime factors are 2 and 3.

### Step 3: Find the Lowest Powers
Now, we take the lowest power of each common prime factor:

- For 2:
- The powers are \(2^4\) (from 48), \(2^3\) (from 72), and \(2^2\) (from 84).
- The lowest power is \(2^2\).

- For 3:
- The powers are \(3^1\) (from 48), \(3^2\) (from 72), and \(3^1\) (from 84).
- The lowest power is \(3^1\).

### Step 4: Calculate the HCF
Now we can calculate the HCF by multiplying the lowest powers of the common prime factors:

\[ \text{HCF} = 2^2 \times 3^1 = 4 \times 3 = 12 \]

### Step 5: Conclusion
Thus, the minimum number of rows required to arrange the balls is **12**.

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