Three pieces of timber 84 m, 98 m and 126 m long have to be divided into planks of the same length. What is the greatest possible length of each plank?

Options

A. 14 m
B. 28 m
C. 7 m
D. 21 m

Answer

To find the greatest possible length of each plank that can be cut from the three pieces of timber measuring 84 m, 98 m, and 126 m, we need to determine the highest common factor (HCF) of these three lengths. Here’s how to do it step by step:

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

1. **For 84 m:**
- Divide by 2: \( 84 \div 2 = 42 \)
- Divide by 2 again: \( 42 \div 2 = 21 \)
- Divide by 3: \( 21 \div 3 = 7 \)
- Divide by 7: \( 7 \div 7 = 1 \)
- So, the prime factorization of 84 is:
\[ 84 = 2^2 \times 3^1 \times 7^1 \]

2. **For 98 m:**
- Divide by 2: \( 98 \div 2 = 49 \)
- Divide by 7: \( 49 \div 7 = 7 \)
- Divide by 7 again: \( 7 \div 7 = 1 \)
- So, the prime factorization of 98 is:
\[ 98 = 2^1 \times 7^2 \]

3. **For 126 m:**
- Divide by 2: \( 126 \div 2 = 63 \)
- Divide by 3: \( 63 \div 3 = 21 \)
- Divide by 3 again: \( 21 \div 3 = 7 \)
- Divide by 7: \( 7 \div 7 = 1 \)
- So, the prime factorization of 126 is:
\[ 126 = 2^1 \times 3^2 \times 7^1 \]

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

- **Common prime factors:**
- The factor 2 appears in all three numbers: \( 2^1 \) (the lowest power).
- The factor 3 appears only in 84 and 126, not in 98.
- The factor 7 appears in all three numbers: \( 7^1 \) (the lowest power).

### Step 3: Calculate the HCF
To find the HCF, we take the lowest power of all common prime factors:

\[ HCF = 2^1 \times 7^1 = 2 \times 7 = 14 \]

### Conclusion
The greatest possible length of each plank that can be cut from the three pieces of timber is **14 meters**.

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