Question

Find the largest number which divides $615$and $963$ leaves the remainder $6$ in each.

Answer 1

Find the HCF for the given set of numbers.

Let $x$ be the required largest number

Subtract$6$ from $615$ and $963$

$$\begin{gathered} 615-6=609 \\ 963-6=957 \end{gathered}$$

Finding HCF using prime factorization,

$$\begin{gathered} 609=29\times 3\times 7 \\ 957=29\times 11\times 3 \end{gathered}$$

The common factors are $29$ and $3$

$$\begin{gathered} \Rightarrow x=29\times 3 \\ \Rightarrow x=87 \end{gathered}$$

Hence, $87$ is the largest number that divides both $615$and $963$ and leaves a remainder $6$ in each case.