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

We need to find the largest number which divides 615 and 963 leaving remainder 6 in each case.

The required number when divides 615 and 963 , leaves remainder 6 , this means $615-6=609$ and $963-6=957$ are completely divisible by the number.

Therefore,

The required number = H.C.F. of 609 and 957.

By applying Euclid’s division lemma

$957=609 \times 1+348$

$609=348 \times 1+261$

$348=216 \times 1+87$

$261=87 \times 3+0$

Therefore, H.C.F. = 87.

Hence, the required number is 87 .