Question

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

Answer 1

Answer = 87

615 -6 = 609

609 = 3 × 7 × 29

963 - 6 = 957

957 = 3 × 11 × 29

609 = 3 × 7 × 29

957 = 3 × 11 × 29

3 × 29 is common 3 × 29 = 87

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.

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