Question

What is the largest number which divides $206,368$ and $449$ and leaves the same remainder in each case?

• A. $32$
• B. $48$
• C. $81$
• D. $96$

Answer 1

The correct option is C $81$

Let the remainder in each case be $r$.

So, $206-r$, $368-r$ and $449-r$ are all divisible by $d$ which is the H.C.F. of these numbers.

Theory: H.C.F. of $a,b,c=$HCF of $(b-a),(c-b)$

Using this theory,

HCF of $\left(\right(368-r)-(206-r\left)\right)$ and $\left(\right(449-r)-(368-r\left)\right)=$

HCF of $(168,81)$

$=81$