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

Solution

It is given that required number when divides 615 and 963, the remainder is 6 in each case.

⇒ 615 - 6 = 609 and 963 - 6 = 957 are completely divisible by the required number.

Since, it is given that the required number is the largest number. Given integers are 957 and 609 clearly 957 > 609.

Therefore, it is the HCF of 609 and 957.

Now, finding HCF by using Euclid’s division lemma to 609 and 957, we get

II. Since the remainder 348 ≠ 0, we apply division lemma to 348 and 609 to get

III. We consider the new divisor 348 and new remainder 261 and apply division lemma to get

IV. We consider the new divisor 261 and new remainder 87 and apply division lemma to get

The remainder at this step is zero. So, the divisor at this stage or the remainder at the previous stage i.e., 87 is the HCF of 615 and 963.

Some More Questions From Real Numbers Chapter

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Real Numbers

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

Some More Questions From Real Numbers Chapter

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