Question

Find the largest number that divides $220,313$,and $716$ leaving remainder $3$ in each case.

Answer 1

We have to find the largest number that divides $220,313$ and $716$ leaving remainder is $3$.

So, first, subtract $3$ from each number

$220-3=217$
$313-3=310$
$716-3=713$
By prime factorizing $217,310$ and $713$ we get
$217=7\times 31$
$310=2\times 5\times 31$
$713=23\times 31$

HCF $=$ Least power of the common prime factor

So, the H.C.F of $217,310$ and $713=31$
Hence the largest number that divides $220,313$ and $716$ leaving a remainder $3$ in each case is $31$.