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

Let the HCF of 615 and 963 be x


Since remainder is 6 in each case, therefore for complete divisible 6 to be subtracted from both the numbers


Therefore new numbers are:


615-6 = 605


963-6 = 957


Prime factors of 609 = 3 × 3 × 29


Prime factors of 957= 3 × 11 × 29


Therefore HCF of 609 and 957 is: 3 × 29 = 87


Hence the largest number which divides 615 and 963 leaving remainder 6 in each case is 87


7