Find the greatest number which divides 615 and 963 , leaving the remainder 6 in each case.

Options

A. 57
B. 67
C. 77
D. 87

Answer

To find the greatest number that divides both 615 and 963, leaving a remainder of 6 in each case, we can follow these steps:

### Step 1: Subtract the remainder from both numbers
To eliminate the remainder, we subtract 6 from both 615 and 963.

\[ 615 - 6 = 609 \]
\[ 963 - 6 = 957 \]

### Step 2: Find the HCF of the two resulting numbers
Next, we need to find the Highest Common Factor (HCF) of 609 and 957. We can do this using prime factorization.

### Step 3: Prime factorization of 609
To factor 609, we can start dividing by the smallest prime numbers:
- 609 is divisible by 3:
\[ 609 \div 3 = 203 \]
- Next, we factor 203. It is also divisible by 7:
\[ 203 \div 7 = 29 \]
- Since 29 is a prime number, we stop here.

So, the prime factorization of 609 is:
\[ 609 = 3 \times 3 \times 29 = 3^2 \times 29 \]

### Step 4: Prime factorization of 957
Now, we factor 957:
- 957 is divisible by 3:
\[ 957 \div 3 = 319 \]
- Next, we factor 319. It is divisible by 11:
\[ 319 \div 11 = 29 \]
- Since 29 is a prime number, we stop here.

So, the prime factorization of 957 is:
\[ 957 = 3 \times 11 \times 29 \]

### Step 5: Identify common factors
Now we identify the common prime factors from both numbers:
- The common prime factors are 3 and 29.

### Step 6: Calculate the HCF
The HCF can be found by multiplying the lowest powers of the common prime factors:
\[ HCF = 3^1 \times 29^1 = 3 \times 29 = 87 \]

### Step 7: Conclusion
The greatest number that divides both 615 and 963, leaving a remainder of 6, is:
\[ \text{Greatest number} = HCF = 87 \]