Find the largest number that divides 189 and 249 leaving remainder 9 in each case .

Answer

To find the largest number that divides both 189 and 249 leaving a remainder of 9, we can follow these steps:

### Step 1: Subtract the remainder from both numbers
Since we want the largest number that divides both 189 and 249 leaving a remainder of 9, we first subtract 9 from both numbers.

- For 189:
\[ 189 - 9 = 180 \]
- For 249:
\[ 249 - 9 = 240 \]

### Step 2: Find the HCF (Highest Common Factor) of the two results
Now we need to find the HCF of 180 and 240.

**Finding the prime factorization of 180:**
1. Divide by 2: \(180 \div 2 = 90\)
2. Divide by 2: \(90 \div 2 = 45\)
3. Divide by 3: \(45 \div 3 = 15\)
4. Divide by 3: \(15 \div 3 = 5\)
5. Divide by 5: \(5 \div 5 = 1\)

So, the prime factorization of 180 is:
\[ 180 = 2^2 \times 3^2 \times 5^1 \]

**Finding the prime factorization of 240:**
1. Divide by 2: \(240 \div 2 = 120\)
2. Divide by 2: \(120 \div 2 = 60\)
3. Divide by 2: \(60 \div 2 = 30\)
4. Divide by 2: \(30 \div 2 = 15\)
5. Divide by 3: \(15 \div 3 = 5\)
6. Divide by 5: \(5 \div 5 = 1\)

So, the prime factorization of 240 is:
\[ 240 = 2^4 \times 3^1 \times 5^1 \]

### Step 3: Identify the common factors
Now, we find the common factors from the prime factorizations:
- For \(2\): The minimum power is \(2^2\)
- For \(3\): The minimum power is \(3^1\)
- For \(5\): The minimum power is \(5^1\)

### Step 4: Calculate the HCF
Now we multiply the common factors:
\[ \text{HCF} = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60 \]

### Conclusion
The largest number that divides both 189 and 249 leaving a remainder of 9 is:
\[ \boxed{60} \]

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