Question

find the largest number which divide 615 and 963 leaving the remainder 6 in each case

Answer 1

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 from both the cases.

609 = 3 x 3 x 29

957= 3 x 11 x 29

⇒ x = 3 x 29 = 87

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

Answer 2

Since remainder in each case is 6
Therefore on subtracting 6 from numbers it will exactly divisible by the largest number
Therefore numbers after subtracting 6 are 609 &957
By HCF we can find the largest number so by Euclid division lemma
957=609×1+348
609=348×1+261
348=261×1+87
261=87×3+0
So 87 is the largest number that divides 957 & 609
Now the numbers are completely divisible by 87 so after taking the real numbers that is 615 & 963
It will leave remainder as 6
Therefore the largest possible no that divides 615 and 96 is 87