Question

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:​

Answer 1

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N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)

  = H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

Answer 2

Answer:

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