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

In order to find the largest number number which divides 615 and 963 leaving remainder 6 in each case, we need to deduct the remainder 6 from both the numbers and find their HCF.

$615-6=609$

$963-6=957$

Prime Factorising both 609 and 657 we get,

$\text{609 = 3 x 3 x 29}$ and

$\text{957= 3 x 11 x 29}$

HCF of $609,957=$ $3\times 29$

$\text{= 3 x 29 = 87}$

∴ The largest number which divides $615$ and $963$ leaving remainder $6$ in each case is $\text{87}$ .