Find the HCF of 2205, 5145 and 4410.

Answer

To find the HCF (Highest Common Factor) of the numbers 2205, 5145, and 4410, we will follow these steps:

### Step 1: Prime Factorization of Each Number

**Prime Factorization of 2205:**
1. Start dividing by the smallest prime number, which is 3.
- 2205 ÷ 3 = 735
2. Divide 735 by 3 again.
- 735 ÷ 3 = 245
3. Now divide 245 by 5.
- 245 ÷ 5 = 49
4. Finally, divide 49 by 7.
- 49 ÷ 7 = 7
5. And divide 7 by 7.
- 7 ÷ 7 = 1

So, the prime factorization of 2205 is:
\[ 2205 = 3^2 \times 5^1 \times 7^2 \]

**Prime Factorization of 5145:**
1. Start dividing by 3.
- 5145 ÷ 3 = 1715
2. Now divide 1715 by 5.
- 1715 ÷ 5 = 343
3. Finally, divide 343 by 7.
- 343 ÷ 7 = 49
4. And divide 49 by 7.
- 49 ÷ 7 = 7
5. And divide 7 by 7.
- 7 ÷ 7 = 1

So, the prime factorization of 5145 is:
\[ 5145 = 3^1 \times 5^1 \times 7^3 \]

**Prime Factorization of 4410:**
1. Start dividing by 2.
- 4410 ÷ 2 = 2205
2. Now we already know the prime factorization of 2205 from above.

So, the prime factorization of 4410 is:
\[ 4410 = 2^1 \times 3^2 \times 5^1 \times 7^2 \]

### Step 2: Identify Common Factors
Now we will identify the common prime factors from the factorizations:
- For \(3\): The minimum exponent is \(1\) (from 5145).
- For \(5\): The minimum exponent is \(1\) (common in all).
- For \(7\): The minimum exponent is \(2\) (from 2205 and 4410).

### Step 3: Calculate HCF
Now we multiply the common prime factors raised to their minimum exponents:
\[ \text{HCF} = 3^1 \times 5^1 \times 7^2 \]

Calculating this:
1. \(3^1 = 3\)
2. \(5^1 = 5\)
3. \(7^2 = 49\)

Now, multiply these together:
\[ \text{HCF} = 3 \times 5 \times 49 \]

Calculating step-by-step:
- First, \(3 \times 5 = 15\)
- Then, \(15 \times 49 = 735\)

Thus, the HCF of 2205, 5145, and 4410 is:
\[ \text{HCF} = 735 \]

### Final Answer:
The HCF of 2205, 5145, and 4410 is **735**.

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