Given that the largest number when divides 248 and 1032, the remainder is 8 in each case.248 - 8 = 240 and 1032 - 8 = 1024 are completely divisible by the required number.Therefore, it is the HCF of 240 and 1024.Prime factorization of 240 = 2 * 2 * 2 * 2 * 3 * 5Prime factorization of 1024 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2.HCF(240,1024) = 2 * 2 * 2 * 2                          = 16.Therefore the largest number which divides 248 and 1032 leaving remainder 8 in each case = 16.

Solution:

To solve this problem, we need to find the highest common factor (HCF) of the numbers obtained by subtracting 8 from 248 and 1032, respectively.

Step 1: Subtract 8 from 248 and 1032

248 - 8 = 240
1032 - 8 = 1024

Step 2: Find the HCF of 240 and 1024

To find the HCF of 240 and 1024, we can use the Euclidean algorithm, which involves dividing the larger number by the smaller number and finding the remainder. Then we divide the smaller number by the remainder and find another remainder. We repeat this process until the remainder becomes zero. The last non-zero remainder is the HCF of the two numbers.

We can write this process as follows:

1024 = 4 × 240 + 64
240 = 3 × 64 + 48
64 = 1 × 48 + 16
48 = 3 × 16 + 0

Therefore, the HCF of 240 and 1024 is 16.

Step 3: Add the common remainder to the HCF

Since we subtracted 8 from both numbers, we need to add 8 to the HCF to obtain the final answer.

16 + 8 = 24

Therefore, the largest number that divides 248 and 1032 leaving a remainder of 8 in each case is 24.

Final Answer: 24