Find the largest number which divides 630 and 940 leaving remainders 6 and 4 respectively

Answer

To find the largest number that divides 630 and 940 leaving remainders of 6 and 4 respectively, we can follow these steps:

### Step 1: Adjust the numbers by subtracting the remainders
We start by subtracting the remainders from the original numbers:
- For 630, we subtract 6:
\[ 630 - 6 = 624 \]
- For 940, we subtract 4:
\[ 940 - 4 = 936 \]

### Step 2: Find the greatest common divisor (GCD)
Next, we need to find the greatest common divisor (GCD) of the two adjusted numbers, 624 and 936.

#### Step 2.1: Prime factorization of 624
To find the GCD, we can start with the prime factorization of 624:
- Divide by 2:
\[ 624 \div 2 = 312 \]
- Divide by 2 again:
\[ 312 \div 2 = 156 \]
- Divide by 2 again:
\[ 156 \div 2 = 78 \]
- Divide by 2 again:
\[ 78 \div 2 = 39 \]
- Now, divide by 3:
\[ 39 \div 3 = 13 \]
- 13 is a prime number.

So, the prime factorization of 624 is:
\[ 624 = 2^4 \times 3^1 \times 13^1 \]

#### Step 2.2: Prime factorization of 936
Now, we perform the prime factorization of 936:
- Divide by 2:
\[ 936 \div 2 = 468 \]
- Divide by 2 again:
\[ 468 \div 2 = 234 \]
- Divide by 2 again:
\[ 234 \div 2 = 117 \]
- Now, divide by 3:
\[ 117 \div 3 = 39 \]
- Divide by 3 again:
\[ 39 \div 3 = 13 \]
- 13 is a prime number.

So, the prime factorization of 936 is:
\[ 936 = 2^3 \times 3^2 \times 13^1 \]

### Step 3: Determine the GCD
Now, we find the GCD by taking the lowest power of each common prime factor:
- For 2: minimum of \(4\) (from 624) and \(3\) (from 936) is \(2^3\)
- For 3: minimum of \(1\) (from 624) and \(2\) (from 936) is \(3^1\)
- For 13: minimum of \(1\) (from both) is \(13^1\)

Thus, the GCD is:
\[ GCD = 2^3 \times 3^1 \times 13^1 = 8 \times 3 \times 13 \]
Calculating this:
\[ 8 \times 3 = 24 \]
\[ 24 \times 13 = 312 \]

### Final Answer
The largest number that divides 630 and 940 leaving remainders of 6 and 4 respectively is:
\[ \boxed{312} \]