What will be the highest number which when divides 52 and 104 leaves remainder 4 and 8 respectively?

Options

A. 12
B. 48
C. 16
D. 26

Answer

To solve the problem, we need to find the highest number that divides both 52 and 104, leaving remainders of 4 and 8 respectively. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We need to find a number \( x \) such that:
- When 52 is divided by \( x \), the remainder is 4.
- When 104 is divided by \( x \), the remainder is 8.

### Step 2: Rewrite the Conditions
From the conditions given, we can rewrite the equations:
- For 52: \( 52 \equiv 4 \mod x \) implies \( 52 - 4 = 48 \) is divisible by \( x \).
- For 104: \( 104 \equiv 8 \mod x \) implies \( 104 - 8 = 96 \) is divisible by \( x \).

### Step 3: Find the Numbers to Consider
Now, we need to find the highest number \( x \) that divides both 48 and 96.

### Step 4: Calculate the HCF (Highest Common Factor)
To find the highest number that divides both 48 and 96, we calculate the HCF of these two numbers.

1. **Prime Factorization of 48**:
- \( 48 = 2^4 \times 3^1 \)

2. **Prime Factorization of 96**:
- \( 96 = 2^5 \times 3^1 \)

### Step 5: Identify Common Factors
To find the HCF, we take the lowest power of each common prime factor:
- For \( 2 \): The lowest power is \( 2^4 \).
- For \( 3 \): The lowest power is \( 3^1 \).

### Step 6: Calculate the HCF
Now, we multiply the common factors:
\[ \text{HCF} = 2^4 \times 3^1 = 16 \times 3 = 48 \]

### Step 7: Conclusion
The highest number \( x \) that divides both 52 and 104, leaving remainders of 4 and 8 respectively, is **48**.