The numbers 2606, 1022 and 4814 when divided by a number N, give the same remainder of 14. Find the highest such number N.

Answer

To find the highest number \( N \) such that the numbers 2606, 1022, and 4814 give the same remainder of 14 when divided by \( N \), we can follow these steps:

### Step 1: Subtract the common remainder from each number
We start by subtracting the common remainder (14) from each of the numbers:

- \( 2606 - 14 = 2592 \)
- \( 1022 - 14 = 1008 \)
- \( 4814 - 14 = 4800 \)

### Step 2: Find the highest common factor (HCF) of the resulting numbers
Next, we need to find the HCF of the three numbers obtained in the previous step: 2592, 1008, and 4800.

### Step 3: Factor each number
Let's find the prime factorization of each number:

1. **For 2592**:
- \( 2592 = 2^5 \times 3^4 \)

2. **For 1008**:
- \( 1008 = 2^4 \times 3^2 \times 7 \)

3. **For 4800**:
- \( 4800 = 2^6 \times 3^1 \times 5^2 \)

### Step 4: Identify the common factors
Now we find the lowest power of each prime factor that appears in all three factorizations:

- For \( 2 \): The minimum power is \( 2^4 \) (from 1008).
- For \( 3 \): The minimum power is \( 3^1 \) (from 4800).
- The primes \( 5 \) and \( 7 \) do not appear in all three numbers, so they are not included.

### Step 5: Calculate the HCF
Now we can calculate the HCF:

\[ \text{HCF} = 2^4 \times 3^1 = 16 \times 3 = 48 \]

### Conclusion
Thus, the highest number \( N \) that gives the same remainder of 14 when dividing 2606, 1022, and 4814 is:

\[ \boxed{48} \]