A number when successively divided by 5, 3, 2 gives remainder 0, 2, 1 respectively in that order. What will be the remainder when the same number is divided successively by 2, 3, 5 in that order

• 2x+1
  3*(2x+1)+2
  5*[3*(2x+1)+2]
  =30x+25
  {30x+25}/2=15x+12(rem=1); [15x+12]/3=5x+4 (rem=0); (5x+4)/5=x (rem=4)
  ans is 1,0,4
• 11 years agoHelpfull: Yes(94) No(4)
• this is the short trick which i am telling you .
  it is applicable only on when the question say that find the remainder when order of divisibility is reversed.
  
  write
  5 0
  3 2
  2 1
  leave the last 2, and multiply 1 by 3 then add 2 after this result will be multiplied by 5 and then add 0 so the remainder no. is 25
  now divide 25 by 2 remainder will be 1 then by 3 rem. will be 0 and then by 5 finally rem. will be 4.
  so 1,0,4 is the right answer.
• 11 years agoHelpfull: Yes(23) No(5)
• Let the number be 55
  So 55%5 = 0 and 55/5 = 11 ; 11%3 = 2 and 11/3 = 3; 3%2 = 1
  According to question,
  55%2 = 1 and 55/2 = 27 ; 27%3 = 0 and 27/3 = 9; 9%5 = 4
  So the remainder will be 1,0,4
• 11 years agoHelpfull: Yes(21) No(8)
• answer is 1,2,0
  the number is 5
  if 5 is divided by 5 it ll gives rem zero
  if 5 is divided by 3 it ll gives rem two
  if 5 is divided by 2 it ll gives rem one
  from the above the number 5 is satisfying the given condition
  so that if 5 is divided successively with 2,3,5, than it ll gives 1,2,0
• 11 years agoHelpfull: Yes(7) No(13)
• as number is divisible by 5 leving 0 and 3 leaving 2 and 2 leaving 1.
  thus number can be written as
  5(3(2m +1) +2)=original number
  simplifying it
  30m +25=original number.
  thus deviding by 2 gives 1 leaving 15m+12
  thus deviding by 3 gives 0 leaving 5m+4
  thus deviding by 5 gives 4
• 11 years agoHelpfull: Yes(7) No(1)
• The no will be ((1*3)+2)5=25 (smallest possible no)
  
  on successive division it will leave 1,0,4 as remainder
• 11 years agoHelpfull: Yes(5) No(3)
• 55 is correct answer... 100% sure
• 11 years agoHelpfull: Yes(3) No(8)
• 35 will be the no
  and rem=1,2,0
  chek out
• 11 years agoHelpfull: Yes(1) No(5)
• n=5x+0
  x=3y+2
  z=2y+1
  then if y=1 the no is div by 2 with rem 1 so that x=5 , n=25
  25/2 rem =1, 12/3 rem=0, 4/5 rem=4
  ans is 1,0, 4
• 11 years agoHelpfull: Yes(1) No(1)
• given 5%x=0 ,3%a=3, 2%b=1
  
• 11 years agoHelpfull: Yes(0) No(0)
• THIS CAN BE SOLVED BY DOING THIS
  First subtracting dividends by reminders respectively then we get
  (5-0),(3-2),(2-1)
  5,1,1
  now LCM(5,1,1)=5
  SO the required number is 5 (check with 5,3,2 we get reminders 0,2,1)
  so 5 is successively divided by 2,3,5 we get the reminders 1,2,0 respectively.
• 11 years agoHelpfull: Yes(0) No(1)
• 5 3 2
  0 2 1
  first multiply 3 to 1 then add 1..after getting the result multiply with 5 and 0
  i.e 3*1+2=5*5+0=25
  then divide 25 by with 2, 3, 5 successively we get 1, 0, 4 as remainder
  so ans is 1, 0, 4
• 10 years agoHelpfull: Yes(0) No(0)
• 1,0,2 . The num is 35
• 6 years agoHelpfull: Yes(0) No(0)
• Can't say, don't know
• 6 years agoHelpfull: Yes(0) No(3)
• Write the divisors and remainders as given below
  
  then solve it as follows
  (((3
  ×
  3) + 2) 2 + 1) = 23
  or 3
  ×
  3 = 9
  9 + 2 = 11
  11
  ×
  2 = 22
  22 + 1 = 23
  So, the least possible number is 23 and the higher numbers can be obtained as (2
  ×
  3
  ×
  5)m + 23 = 30m + 23.
  So the higher numbers are 53, 83, 113, 143, 173, 213, 333, . .. . etc.
  But when we divide all these possible numbers by 7 we get the different remainders.
• 5 Months agoHelpfull: Yes(0) No(0)