A number when divided by D leaves a remainder of 8 and when divided by 3D leaves a remainder of 21. What is the remainder left, when twice the number is divided by 3D?
a) 13
b) cannot be determined
c) 3
d) 42

• If Number is N and Divisor is D then from given conditions
  N = a*D + 8 --- 1
  N = b*3D + 21 --- 2
  
  We have to find R
  2N = c*3D + R
  
  from (2)- (1) we get
  D(3b - a) = 21 - 8 = 13
  
  This implies D is a factor of 13
  now D cannot be one therefore D=13
  
  From 2
  2N = 2b*3D + 21*2 --- (2)*2
  = K*3D + 42
  = K*3*(13) + 13*3 + 3 --as D=13
  = (K + 1)3*13 + 3 -- Rearranging the Divisor part to insert 3*13
  = c*3D + 3
  
  Thus 2N = c*3D + 3
  Hence remainder will be 3
  
  For basic concept explanation please find a more complex sample with solution details
  If there is Number N and Divisor D such that
  N=aD + 17 -- 1
  3N = bD +7 --2
  What is R in below equation?
  8N = cD + R
  Solution:
  Now from 3(1) - (2)
  D(3a-b) = 17*3 - 7 = 44
  ==> D = 2,4,11,22,44
  Remainder is 17 when N is divided by D
  hence D cannot be less than 17
  ==> D= 22 or 44
  Now
  8N = 8aD + 8*17 -- from 8*(1)
  = 8aD + 136
  
  8aD is divisible by D so eliminated
  Focus on 136 when divided by D (22 or 44)
  Remainder(136/22) = Remainder(136/44) = 4
  Hence
  8N = 8aD + 4
  OR
  8N = cD + 4
  
  8N = (8a + 3)D + 4 // In case of D=44
  8N = (8a + 6)D + 4 // In case of D=22
  This is basic funda
  if case where D > 136 then remainder wud be 136 only
  if case where D < 136 like here then aproach like above
  --Amit Pratap Singh
• 13 years agoHelpfull: Yes(5) No(10)
• answer will be 15
  
• 13 years agoHelpfull: Yes(0) No(4)