Question

find the largest number that divides 615and963leaving remainder 6 in each case​

Answer 1

Given :

• 615 and 963.

• Remainder = 6 in each case.

To find :

• The largest number that divides 615 and9 63 leaving remainder 6 in each case.

Step-by-step explanation :

It is Given that,

615 and 963 leaving remainder 6 in each case.

So, we have to reduce 6 from both divisors.

615 - 6 = 609

963 - 6 = 957

Now,

We have to find the largest number that divides 615 and 963 leaving remainder 6 in each case. So, firstly find the HCF (Highest Common Factor) of both divisors.

HCF of 609 and 957 :-

609 = 3 × 3 × 29

957 = 3 × 11 × 29

•°• HCF (Highest Common Factor) = 3 × 29 = 87

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

Answer 2

$\bigstar$We have to find the largest number which divide 615 and 963 leaving the remainder 6 in case.

So, let us subtract 6 from both 615 and 963.

$\longrightarrow \sf\:615-6=609 \\ \\ \longrightarrow \sf\:963-6=957$

Now let us find the HCF of 609 and 957

$\bigstar$ Prime factorisation of 609

$\longrightarrow$29 × 3 × 3

$\bigstar$ Prime factorisation of 957

$\longrightarrow$29 × 3 × 11

Now, let us take out the common factors from both numbers :

$\longrightarrow \sf\: 29\:and\:3$

HCF = 29 × 3 = 87

Therefore, $\boxed{87}$ is the number which will divide 615 and 963 leaving remainder 6 in each case.