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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.
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.
Therefore, the largest number that will divide 615 and 963 leaving remainder 6 in each case is 87.
Updated on: 10-Oct-2022
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