Find the largest number which divides 615 and 963 leaving remainder 6 in each case.

Hint: We have to find the largest factor of 615 and 963 such that the remainder is 6 for both the given numbers. So, we will find the highest common factor of $615-6=609$ and $963-6=957$. We will find the factors of 609 and 957. Next, we will group together the common factors of both the given numbers and then find the highest common factor.

Complete step by step answer:
To find the largest number which divides 615 and 963 such that it leaves a remainder of 6 in each case is the same as finding the highest common factor of $615-6=609$ and $963-6=957$.
First, we will factor the number 609. The first prime that divides 609 is 3. So we factor 609 by 3 in the following manner,
\[609=3\times 203\]
Now, the first prime that divides 203 is 7. So, factoring 203 by 7, we get
$609=3\times 7\times 29$
We have factored 609 by 3, 7 and 29. Further factorization is not possible as 3, 7 and 29 are all prime numbers.
Now, we will factor the number 957. The first prime that divides 957 is 3. Factoring 957 by 3, we get
$957=3\times 319$
Next, we will factor the number 319. The number 319 is divisible by 11. Therefore, we get the following factorization,
$957=3\times 11\times 29$
The number 957 has factors 3, 11 and 29. There cannot be factors more than these three, as these three are prime numbers.
Now, we can see that 3 and 29 are the common factors of 609 and 957. Therefore, the highest common factor of 609 and 957 is $29\times 3=87$. This means, 87 is the largest number which divides 615 and 963 leaving remainder 6 in each case.

So, the correct answer is 87.

Note: The numbers are large and it is difficult to come up with factors for these numbers in one glance. Hence, it is useful to know some tests of divisibility for this question. We should be well versed with the concept of highest common factor. Also, subtracting the remainder 6 beforehand is not a necessity. We can factorize 615 and 963 in a way that leaves 6 as a remainder.