Question

find the largest positive integer which divides 615 and 963 leaving remainder 6 in each case​

Answer 1

Answer:

87

To find the largest number which divides 615 and 963 leaving remainder 6 in each case i.e. HCF. In order to make 615 and 963 completely divisible by x, we need to deduct the remainder 6 from both the cases. ∴ largest number which divides 615 and 963 leaving remainder 6 in each case is 87.

Step-by-step explanation:

hope its helps

Answer 2

$\huge\fbox \red{Solution:}$

Firstly, the required numbers which on dividing doesn’t leave any remainder are to be found.

This is done by subtracting 6 from both the given numbers.

So, the numbers are 615 – 6 = 609 and 963 – 6 = 957.

Now, if the HCF of 609 and 957 is found, that will be the required number.

$\fbox \orange{By applying Euclid’s division lemma}$

957 = 609 x 1+ 348

609 = 348 x 1 + 261

348 = 261 x 1 + 87

261 = 87 x 3 + 0.

⇒ H.C.F. = 87.

$\fbox \blue{Therefore, the required number is 87.}$