When 100 is to be successively divided by 6, 3, 4, first divide 100 by 6. Then divide the quotient 16 by 3. Then divide the quotient 5 by 4. A number when successively divided by 5, 3, 2 gives the remainder of 0, 2 and 1 respectively in that order. What will be the remainders when the same number is divided successively by 2, 3 and 5 in that order?
1) 1, 0, 4 2) 4, 3, 2 3) 4, 1, 2 4) 2, 1, 3

• 1) 1, 0, 4
  Let the number be A
  So, when divided by 5 : A = 5K + 0 = 5K
  Thus, when you divide A by 5 you get K
  According to question, the number is successively then divided by 3.
  
  From above when A is divided by 5. We get K
  So, dividing successively, K is divided further by 3 and leaves remainder 2
  So, following is valid : K = 3J + 2
  Again, according to the question, the number is again successively divided by 2
  
  From the previous step we get J, after K is divided by 3.
  Thus, dividing successively, J is divided further by 2 and leaves remainder 1
  So, following is valid : J = 2L + 1
  We get the following three equations:
  
  J = 2L + 1
  K = 3J + 2
  A = 5K
  Replacing values in A = 5K equation
  
  A = 5K
  A = 5(3J + 2) = 15J + 10
  A = 15(2L + 1) + 10 = 30L + 25
  Any number of the above conditions will satisfy the following equation -
  
  Num : 30L + 25 (Where L can be 0, 1 , 2, 3 ......... )
  If L = 0 then, number is 25
  
  Dividing 25 by 2 we get
  12 and 1 remainder
  Dividing 12 by 3 we get (Since, successively)
  4 and 0 remainder
  Dividing 4 by 5 we get (Since, successively)
  0 and 4 remainder
  So answer is : 1, 0, 4
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