Three pieces of timber of lengths 63m, 42m and 35m, 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 pieces of timber measuring 63m, 42m, and 35m, we need to determine the highest common factor (HCF) of these three lengths. Here’s how to do it step by step:

### Step 1: Find the prime factorization of each length.
- **For 63**:
- 63 can be divided by 3:
- \( 63 \div 3 = 21 \)
- 21 can be divided by 3:
- \( 21 \div 3 = 7 \)
- So, the prime factorization of 63 is \( 3^2 \times 7 \).

- **For 42**:
- 42 can be divided by 2:
- \( 42 \div 2 = 21 \)
- 21 can be divided by 3:
- \( 21 \div 3 = 7 \)
- So, the prime factorization of 42 is \( 2^1 \times 3^1 \times 7^1 \).

- **For 35**:
- 35 can be divided by 5:
- \( 35 \div 5 = 7 \)
- So, the prime factorization of 35 is \( 5^1 \times 7^1 \).

### Step 2: Identify the common factors.
Now, we will list the prime factors of each number:
- 63: \( 3^2, 7^1 \)
- 42: \( 2^1, 3^1, 7^1 \)
- 35: \( 5^1, 7^1 \)

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

### Step 3: Determine the HCF.
The highest common factor (HCF) is the product of the lowest powers of all common prime factors:
- The only common prime factor is \( 7^1 \).
- Therefore, the HCF is \( 7 \).

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

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