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 three pieces of timber measuring 63m, 42m, and 35m, we need to determine the highest common factor (HCF) of these three lengths. Here are the steps to solve the problem:

### Step 1: Factorize each length
- **63**: The prime factorization of 63 is \(3 \times 3 \times 7\) or \(3^2 \times 7\).
- **42**: The prime factorization of 42 is \(2 \times 3 \times 7\).
- **35**: The prime factorization of 35 is \(5 \times 7\).

### Step 2: Identify the common factors
Now, we will identify the common prime factors from the factorizations:
- From 63: \(3^2, 7\)
- From 42: \(2, 3, 7\)
- From 35: \(5, 7\)

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

### Step 3: Determine the HCF
The HCF is the product of the lowest powers of all common prime factors. Here, the only common factor is \(7\) (which has the power of \(1\) in all factorizations). Therefore, the HCF of 63, 42, and 35 is:
\[ \text{HCF} = 7 \]

### Step 4: Conclusion
The greatest possible length of each plank that can be cut from the three pieces of timber is **7 meters**.