Introduction:
In this problem, we are asked to find the greatest number that will divide 148, 246, and 623 giving remainder 4, 6, and 11 respectively.

Solution:
To solve this problem, we can use the method of finding the highest common factor (HCF) of the given numbers.

Step 1:
We can begin by finding the HCF of the given numbers 148, 246, and 623.

Step 2:
Next, we can consider the remainders 4, 6, and 11. We know that if a number divides another number leaving a remainder r, then it will also divide the difference between the two numbers leaving the same remainder. For example, if a number divides 10 leaving a remainder 3, then it will also divide 17 leaving the same remainder 3.

Step 3:
Using this property, we can subtract the remainders from the corresponding numbers to get new numbers that are divisible by the required number.
- 148 - 4 = 144
- 246 - 6 = 240
- 623 - 11 = 612

Step 4:
Now, we can find the HCF of these new numbers, which will be the required number.
- HCF(144, 240, 612)

Step 5:
We can use the prime factorization method to find the HCF.
- Prime factorization of 144 = 2^4 x 3^2
- Prime factorization of 240 = 2^4 x 3 x 5
- Prime factorization of 612 = 2^2 x 3^2 x 17

Step 6:
To find the HCF, we need to take the product of the common prime factors with the lowest exponent.
- HCF = 2^2 x 3 = 12

Conclusion:
Therefore, the greatest number that will divide 148, 246, and 623 giving remainder 4, 6, and 11 respectively is 12.