Find the largest no. which divides 615 and 963 leaving remainder 6 in

1.	Find the largest no. which divides 615 and 963 leaving remainder 6 in each case.
Answer» Since 6 is the remainder in each case Therefore, 615 - 6=609And,963 - 6=957Now required no.=hcf of 609 and 957 957=609×1+348 609=348×1+261 348=261×1+87 261=87×3+0Therefore, required no.=87 To find the largest number which divides 615 and 963 leaving remainder 6 in each case i.e. HCF.Consider HCF be x.In order to make 615 and 963 completely divisible by x, we need to deduct the remainder 6\xa0from both the cases.609 = 3 x 3 x 29957= 3 x 11 x 29⇒ x = 3 x 29 = 87∴\xa0largest number which divides 615 and 963 leaving remainder 6 in each case is 87.

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