Since the remainder is same in each case, hence we will use the following formula
H.C.F (x, y, z) = H.C.F of (x -y), (y- z), ( z-x)
N = H.C.F. of (6905 - 4665), (4665 - 1305), and (6905 - 1305)
= H.C.F. of 2240, 3360 and 5600 (1 mark)
HCF of 5600 and 3360:
5600=3360×1+2240
3360=2240×1+1120
2240=1120×2+0
∴ HCF of 5600 and 3360 = 1120 (1 mark)
Now, HCF of 2240 and 1120 = 1120
So, the HCF of (3360, 2240 and 5600) = N = 1120
Sum of digits in N = (1 + 1 + 2 + 0) = 4 (1 mark)