a nuber when divided by D give remainder 23 abd when by 12D give 104... what will be remainder when divide by 6D.

• for 12d, N can be written as 12dx+104
  
  for d, N = dy + 23
  
  now N will be the same in both cases.
  
  so, N/d = (12dx+104)/d = (dy+23)/d
  
  or, 12x + 104/d = y + 23/d
  
  or, 12x + 81/d + 23/d = y + 23/d
  
  now LHS has to be equal to RHS...
  
  thr4, d should be a factor of 81
  
  since remainder is 23, d>23 which leaves us with d= 27 or 81
  
  take either d = 81 or d = 27, N can be min 12*81 + 104 or 12*27 + 104................................( x = 1)
  
  or, N = 1076 or N = 428
  
  when we divide n by 81 or 27, we do get rem 23 .
  
  now when we divide N by 6*81 = 486 or 6*27 = 162.....we get remainder 104
• 11 years agoHelpfull: Yes(21) No(4)
• let n is the number when it is divided by d.... n= dy +23
  " " 12d....n= 12dx +104
  6d is a multiple of 12d ..... =2*6dx +104..
  so, 6d is also having the same remainder....
• 11 years agoHelpfull: Yes(3) No(2)
• number be x As dividend=divisor*quot+remainder
  x=Dy+23abd
  x=12Dz+104
  x/6=(12Dz+104)6 thus 104/6=2 as 12 is divisible by 6 any num of form 12*Dz is also divisible by 6 thus will not lead to a remainder.
  
• 11 years agoHelpfull: Yes(1) No(6)
• 104 only.... on such problems take ur own an easy example and test u ll get the result
• 11 years agoHelpfull: Yes(0) No(8)
• Number(N)= D*Q+R
  N=D*Q+23 and N=12mD+104
  ans=104
• 11 years agoHelpfull: Yes(0) No(2)