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