To find:
the largest number which divides 615 and 963 leaving remainder 6 in each case.
Solution:
Let the HCF of 615 and 963 be x.
Since it is given remainder is 6 in each case,
Therefore for the numbers to be completely divisible,
6 should be subtracted from both the numbers.
Therefore new numbers are:
615 - 6 = 609
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.
