Find the greatest number which divides 615 and 963, leaving remainder 6 in each case
Solution:
The largest number which divides 615, 963 and leaves the remainder 6
so, subtract 6 from the given numbers
$615 - 6 = 609$
$963 - 6 = 957$
Now, find H C F of 609 and 957
Prime factorization of 609 =$3 \times 3 \times 29$
Prime factorization of 957 = $3 \times 11 \times 29$
Common factors from the above , $3 \times 29 = 87$
So, 87 is the largest number which divides 615, 963 and leaves the remainder 6
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