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# Find the largest number which divides 615 and 963 leaving remainder 6 in each case.

__Given__:

615 and 963

__To find__:

We have to find the value of the largest number which divides 615 and 963 leaving the remainder 6 in each case.

__
__

__Solution__:

If the required number divide 615 and 963 leaving remainder 6 in each case, then this means that number will divide 609(615 - 6) and 957(963 - 6) completely.

Now, we just have to find the HCF of 609 and 957.

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__

__
__

__HCF of 609 and 957:__

__
__

__
__

- Prime factorization of 609 = 3 $\times $ 7 $\times $ 29

- Prime factorization of 957 = 3 $\times $ 11 $\times $ 29

Product of common prime factors = 3 $\times $ 29 = 87

So, HCF of 609 and 957 is 87.

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**Therefore, the largest number that will divide 615 and 963 leaving remainder 6 in each case is 87.**

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