Question

Find the largest number which divides $245$ and $1029$ leaving remainder $5$ in each case.

Answer 1

It is given that the required number when divides $245$ and $1029$, the remainder is $5$ in each case, This means that $245-5=240$ and $1029-5=1024$ are completely divisible by the required number.

It follows from this that the required number is a common factor of $240$ and $1024$. It is also given that the required number is the largest number satisfying the given property. Therefore, it is the $HCF$ of $240$ and $1024$.

Let us now find the $HCF$ of $240$ and $1024$ by Prime factorization method(refer above image).
Clearly, $HCF$ of $240$ and $1024$ is common divisor i.e., $2\times 2\times 2\times 2=16$.
Hence, required number $=16$.