When a number is successively divided by 5, 3 and 2 leaves a remainder of 0, 2 and 1 respectively. What are the remainders when the same number is divided by 2, 3 and 5 successively?

• 2*x+1
  3(2*x+1)
  5(3(2*x+1))=5(6x+5)=30x+25 it will be the number.
  
  when divided by 2 remainder will be 1, quotient will be 15x+12;
  when divided by 3 remainder will be 0, quotient will be 5x+4;
  when divided by5 remainder will be 4.
  so remainders will be 1,0,4
  
  
  
• 9 years agoHelpfull: Yes(9) No(2)
• z=2*1+1=3, y=3z+2=11, x=5y=55
  55 divide by 2 ;1 ,27 divide by 3;0 ,9 divide by 5 ;4
  
• 9 years agoHelpfull: Yes(6) No(1)
• when divided by 5 we get X=5L
  When divided by 3 we get L=3k+2
  when divided by 2 we get k=2M+1
  NOW X=5*(3(2M+1)+2)=30M+25
  ASSUME M=0
  WHEN 25 IS DIVIDED BY 2 WE GET REMAINDER 1 AND QUOTIENT 12
  WHEN 12 IS DIVIDED BY 3 WE GET 0 REMAINDER AND QUOTIENT 4
  WHEN 4 IS DIVIDED BY 5 WE GET 4
  SO ANS IS 1,0,4
• 9 years agoHelpfull: Yes(5) No(0)
• The remainders will be 1,0,4 respectively...................
• 9 years agoHelpfull: Yes(3) No(0)
• ans: 1,2,0 respectively.
• 9 years agoHelpfull: Yes(1) No(2)
• 5 0
  3 2
  2 1
  
  (1*3+2)=5
  5*5+0=25
  the no is 25
  
• 9 years agoHelpfull: Yes(0) No(1)
• as the no is divided by 5 ,3 and 2 leaves remainder 0,2,1 respectively..
  so the only no that is divided by 5,3,2 leaves remainder is 0,2,1 is 5 ..again 5 is dived by 2,3,5 leaves remainder 1,2,0 ans is 1,2,0
• 9 years agoHelpfull: Yes(0) No(2)
• first take 5 as a divisor ,N as a dividend and Q1 as a quotient
  now from the division meathod we know that
  DIVIDEND =DIVISOR *QUOTIENT +REMINDER
  so now N =5*Q1+0..........(1)
  Similarly
  Q1=3*Q2+2.......(2)
  Q2=2*Q3+1........(3)
  put Q3 as a zero then ...
  eqn 1,2,3 will be
  25,5,1......
  now take 25 as a new DIVIDEND and sole with new numbers
  which are 2,3,5
  so reminder will be 1,0,4.... ans.
  thank you
  
• 9 years agoHelpfull: Yes(0) No(1)
• Sol: By given condition, we have
  A+B+C = 13...(1)
  C+D+E = 13...(2)
  E+F+G = 13...(3)
  G+H+I = 13...(4)
  Add up (1)+(2)+(3)+(4), we get
  52=13*4 = A + B + C+ C + D + E+ E + F + G+ G + H + I = C+E+G+(A + B + C+ D +E + F + G + H + I)
  = C+E+G+(1+2+ 3+ 4+ 5+ 6+ 7+ 8+ 9) = C+E+G+(1+9)*9/2 = C+E+G + 45
  
  Hence,C+E+G = 52 - 45 = 7...(5)
  By (5), we have E = 7-C-G
• 9 years agoHelpfull: Yes(0) No(0)
• let the final quotient be x
  now it was obtained by dividing the dividend by 2 leaving 1 as remainder
  so to find the first dividend the expression looks like
  2*x+1.
  repeating the same process for the 2nd dividend we get 3*(2*x+1)+2.
  and finally the actual number is calculated as 5*(3*(2*x+1)+2).
  This expression can be rearranged to be made to look like 30*x+25.
  
  now successive division of the number by 2,3,5 respectively we get 1,0,4 as remainders which are the required answers
• 9 years agoHelpfull: Yes(0) No(0)