Q. If N is the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case.What is the sum of the digits of N ?

• 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
• 12 years agoHelpfull: Yes(35) No(10)
• N = H.C.F. of (4665-1305)=3360, (6905-4665)=2240 and (6905-1305)=5600
  we can write it as
  3360 = 2*2*2*2*2*3*5*7
  2240 = 2*2*2*2*2*2*5*5*7
  5600 = 2*2*2*2*2*5*5*7
  so, HCF of (3360, 2240 and 5600) = 2*2*2*2*2*5*7 = 1120
  Sum of digits of N = ( 1 + 1 + 2 + 0 ) = 4
• 12 years agoHelpfull: Yes(7) No(0)
• Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
  
  Let 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]
• 12 years agoHelpfull: Yes(3) No(8)
• for this we have to find the H.C.F of numbers:
  so,H.C.F of 1305 and 4665 is 65 and so on complete will be 1120
  then sum of digit is:4
  
• 12 years agoHelpfull: Yes(3) No(6)
• 4665-1305=3360
  6905-1305=5600
  6905-4665=2240
  HCF of 2240,3360,5600=1120
  2240)3360(1
  2240
  ----
  1120)2240(2
  2240
  ----
  0000
  1120)5600(5
  5600
  ----
  0000
  HCF=1120
  ans=1+1+2+0
  =4
  second method of finding HCF
  2240=2*2*2*2*2*2*5*7
  3360=2*2*2*2*2*3*5*7
  560=2*2*2*2*2*5*5*7
  HCF=2*2*2*2*25*7
  =1120
  so sum of digit of N=1+1+2+0=4
• 10 years agoHelpfull: Yes(2) No(1)
• 4665-1305=3360
  6905-4665=2240
  now: 3360-2240=1120
  1+1+2+0=4
• 10 years agoHelpfull: Yes(0) No(0)