when a number divided by D leaves remainder of 8and when by 3D leaves remainder of 21.what is the remainder left,when twice the number diveded by 3D.

a-32
b-4567
c-3

• 3
  
  
  Number is 21 and D is 13.
  when 21 is divided by 13 leaves remainder of 8and when by 39 leaves remainder of 21.
  3 is the remainder left
  42 is divided by 39.
• 12 years agoHelpfull: Yes(11) No(4)
• Let number is n
  and k,l,m are positive integers.
  
  Then:
  n=kd+8 (1)
  n=l(3d) + 21 (2)
  from above two equations:
  (3*l-k)d+13=0
  which means either d|13 or (k-3*l)|13
  but d can't be 1 and 13 is prime so d=13
  
  so,
  n=13k+8, n=39*l+21
  k,l >= 0, n is also integer,
  k=1, l=0 is one solution (For which answer of question is 3)
  and others are: k-3*l=1 or k=3*l+1
  
• 12 years agoHelpfull: Yes(3) No(6)
• suppose the no. is a
  a = Dx +8 ...1
  a = 3Dy + 21 ...2
  So from 1 & 2 we get
  D(x-3y) = 13
  13 is prime so x-3y has to be 1
  let x = 4 and y = 1
  We get a = 60
  2a = 120
  So 120 when divided by 3D = 39
  we get remainder as 3
  So answer is 3
• 12 years agoHelpfull: Yes(3) No(0)
• ans is 3
  
  the number is 60 and D=13
  as 60/13=4 (remainder 8)
  and 60/3*13=60/39=1 (remainder 21)
  now 2x/3D=2*60/3*13=120/39=3 (3 remainder)
• 12 years agoHelpfull: Yes(1) No(0)