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Let $N$ be the greatest number that will divide $1305, 4665$ and $6905$, leaving the same remainder in each case. Then sum of the digits in $N$ is

1. $4$
2. $5$
3. $6$
4. $8$

If $N$ be the greatest number that will divide $a,b\;\text{and}\; c,$ leaving the same remainder in each case.

Then $N = \text{HCF of}\;[(\mid a – b \mid), (\mid b– c \mid), (\mid c– a \mid)]$

Now, $N = \text{HCF of}\;[(\mid 1305 – 4665 \mid),\;(\mid 4665 – 6905 \mid),\;(\mid 6905 – 1305\mid)$

$\implies N = \text{HCF of}\;(3360,2040,5600)$

$\implies N = 1120$

$\therefore$ The sum of the digits in $N$ is $= 1 + 1 + 2 + 0 = 4.$

So, the correct answer is $(A).$
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