What will be the HCF of 1785,1995 and 3381?

Answer

To find the HCF (Highest Common Factor) of the numbers 1785, 1995, and 3381, we will follow these steps:

### Step 1: Factor each number
We will start by finding the prime factorization of each number.

**For 1785:**
1. Divide by 3:
- \(1785 \div 3 = 595\)
2. Divide by 5:
- \(595 \div 5 = 119\)
3. Divide by 7:
- \(119 \div 7 = 17\)
4. 17 is a prime number.

Thus, the prime factorization of 1785 is:
\[ 1785 = 3^1 \times 5^1 \times 7^1 \times 17^1 \]

**For 1995:**
1. Divide by 3:
- \(1995 \div 3 = 665\)
2. Divide by 5:
- \(665 \div 5 = 133\)
3. Divide by 7:
- \(133 \div 7 = 19\)
4. 19 is a prime number.

Thus, the prime factorization of 1995 is:
\[ 1995 = 3^1 \times 5^1 \times 7^0 \times 19^1 \]

**For 3381:**
1. Divide by 3:
- \(3381 \div 3 = 1127\)
2. 1127 is not divisible by 3, 5, or 7. Next, we check for 23:
- \(1127 \div 23 = 49\)
3. 49 is \(7^2\).

Thus, the prime factorization of 3381 is:
\[ 3381 = 3^1 \times 7^2 \times 23^1 \]

### Step 2: Identify common factors
Now, we will identify the common prime factors from the factorizations:
- **1785:** \(3^1 \times 5^1 \times 7^1 \times 17^1\)
- **1995:** \(3^1 \times 5^1 \times 7^0 \times 19^1\)
- **3381:** \(3^1 \times 7^2 \times 23^1\)

The common prime factors are:
- \(3^1\) (present in all three numbers)
- \(7^0\) (not present in 1995, so we ignore it)

### Step 3: Calculate the HCF
The HCF is the product of the lowest powers of all common prime factors:
- From the common factors, we only have \(3^1\).

Thus, the HCF is:
\[ HCF = 3^1 = 3 \]

### Final Answer
The HCF of 1785, 1995, and 3381 is **3**.

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