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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Updated on: 10-Oct-2022

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