Three pieces of timber, 42-m, 49-m and 63-m long, have to be divided into planks of the same length. What is the greatest possible length of each plank

Answer

To find the greatest possible length of each plank that can be cut from the three pieces of timber measuring 42 meters, 49 meters, and 63 meters, 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
We will start by finding the prime factors of each of the lengths.

- **For 42:**
- 42 can be divided by 2 (the smallest prime number):
- \( 42 ÷ 2 = 21 \)
- Next, we factor 21:
- 21 can be divided by 3:
- \( 21 ÷ 3 = 7 \)
- So, the prime factorization of 42 is:
- \( 42 = 2 × 3 × 7 \)

- **For 49:**
- 49 is \( 7 × 7 \) or \( 7^2 \).
- So, the prime factorization of 49 is:
- \( 49 = 7 × 7 \)

- **For 63:**
- 63 can be divided by 3:
- \( 63 ÷ 3 = 21 \)
- We already factored 21:
- \( 21 = 3 × 7 \)
- So, the prime factorization of 63 is:
- \( 63 = 3 × 3 × 7 \) or \( 3^2 × 7 \)

### Step 2: Identify Common Factors
Now, we will list the prime factors we found:

- **42:** \( 2, 3, 7 \)
- **49:** \( 7, 7 \)
- **63:** \( 3, 3, 7 \)

The common prime factor among all three numbers is \( 7 \).

### Step 3: Determine the HCF
The highest common factor (HCF) is the product of the common prime factors. Since the only common prime factor is \( 7 \):

- **HCF of 42, 49, and 63 is:** \( 7 \)

### Conclusion
Thus, the greatest possible length of each plank that can be cut from the timber is **7 meters**.

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