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