What is the greatest number which will divide 110 and 128 leaving a remainder 2 in each case?

Options

A. 8
B. 18
C. 28
D. 38

Answer

To find the greatest number that will divide both 110 and 128 leaving a remainder of 2 in each case, we can follow these steps:

### Step 1: Adjust the numbers by subtracting the remainder
Since we want the greatest number that divides 110 and 128 leaving a remainder of 2, we first subtract 2 from both numbers.

- For 110:
\[ 110 - 2 = 108 \]
- For 128:
\[ 128 - 2 = 126 \]

### Step 2: Find the HCF (Highest Common Factor) of the adjusted numbers
Now, we need to find the HCF of 108 and 126.

### Step 3: Prime factorization of 108
Let's perform the prime factorization of 108:
- Divide by 2:
\[ 108 \div 2 = 54 \]
- Divide by 2 again:
\[ 54 \div 2 = 27 \]
- Divide by 3:
\[ 27 \div 3 = 9 \]
- Divide by 3 again:
\[ 9 \div 3 = 3 \]
- Finally, divide by 3:
\[ 3 \div 3 = 1 \]
Thus, the prime factorization of 108 is:
\[ 108 = 2^2 \times 3^3 \]

### Step 4: Prime factorization of 126
Now, let's perform the prime factorization of 126:
- Divide by 2:
\[ 126 \div 2 = 63 \]
- Divide by 3:
\[ 63 \div 3 = 21 \]
- Divide by 3 again:
\[ 21 \div 3 = 7 \]
- Finally, divide by 7:
\[ 7 \div 7 = 1 \]
Thus, the prime factorization of 126 is:
\[ 126 = 2^1 \times 3^2 \times 7^1 \]

### Step 5: Identify the common factors
Now, we identify the common prime factors between 108 and 126:
- The common prime factors are \(2\) and \(3\).
- The minimum power of \(2\) in both factorizations is \(2^1\).
- The minimum power of \(3\) in both factorizations is \(3^2\).

### Step 6: Calculate the HCF
Now, we can calculate the HCF:
\[ HCF = 2^1 \times 3^2 = 2 \times 9 = 18 \]

### Conclusion
The greatest number that will divide both 110 and 128 leaving a remainder of 2 is:
\[ \boxed{18} \]