Question

let N be the greatest number that will divide 1305, 4665 and 6905 leaving the same remainder In each case then sum of digits in N is

Answer 1

Since the remainder is same in each case, we will use the following formula-
 
H.C.F (x, y, z)= H.C.F of (x -y), (y- z), ( z-x)

So, let's take out the H.C.F of 1305, 4665 and 6905 (leaving the same remainder) = H.C.F of (4665 - 1305), ( 6905- 4665) and ( 6905- 1305) 

=H.c.f of 3360, 2240 and 5600 which is= 1120.

Sum of digits of 1120= (1+1+2+0)= 4.

Answer=4.

Answer 2

Answer:

4

Step-by-step explanation:

4665−1305=3360,6905−4665=2240,

6905−1305=5600

Required number (N) = HCF of 3360, 2240 and 5600

⇒N=1120

⇒ Sum of digits of N=1+1+2+0=4