A number when divided by 'D' leaves remainder of 8 and when divided by 3D leaves a remainder of 21.what is the remainder left,when twice of the number is divided by 3D.
a-13
b-can not be determined
c-3
d-42

• suppose the number is N and the quotient is Q and Q'.
  now, N=DQ+8 ..(1)
  N=3DQ'+21 ...(2)
  
  now as we multiply the (1) by 3 and subtract it from (2) we would get
  2N=3D(Q-Q')+3
  hence the remainder is 3.
• 13 years agoHelpfull: Yes(13) No(0)
• Let the number be n. Then there are integers p and q such that
  n = pD + 8 (So 8 < D)...............(1)
  and
  n = 3qD + 21 (So 21 < 3D)........(2)
  
  Thus
  pD + 8 = 3qD + 21
  (p - 3q)D = 13
  
  Now since 13 is a prime, has just two divisors - namely 1 and 13 itself.
  But from above, D > 1 which implies that p-3q = 1
  Hence D = 13
  which, from Eq(2) gives
  n = 39q + 21
  and
  2n = 39*(2q) + 42
  = 39*(2q + 1) + 3
  
  So 2n has remainder 3 when divided by 3D = 39.
• 13 years agoHelpfull: Yes(3) No(0)