find 6 digit number that can be formed by 1,2,3,4,5,6 once such that the 6 digit number is divisible by its unit digit(the unit digit is not 1)

• Since, there is no repetition, numbers 1 to 6 will always be present in the 6 digit number.
  We will go through this step by step.
  
  unit digit divisibility by 2 and 2 at the unit place
  _ _ _ _ _ 2= 5*4*3*2*1*1= 120 ways
  
  unit digit divisibility by 3 and 3 at the unit place
  
  _ _ _ _ _3=5*4*3*2*1*1 =120 ways
  (1+2+3+4+5+6 =21 divisible by 3)
  
  
  unit digit divisibility by 4 and 4 at the unit place
  
  _ _ _ _ 2 4 =4*3*2*1*1 =24
  
  _ _ _ _ 6 4 = 4*3*2*1*1=24
  
  so total ways =24+24 = 48 ways
  
  unit digit divisibility by 5 and 5 at the unit place
  
  _ _ _ _ _ 5 = 5*4*3*2*1*1 = 120 ways
  
  unit digit divisibility by 6 and 6 at the unit place
  
  _ _ _ _ _6=5*4*3*2*1*1= 120ways
  
  total ways =120 +120+ 48+ 120 + 120 = 528 ways
• 7 years agoHelpfull: Yes(17) No(7)
• Something missing in this question it should be how many 6 digit numbers....
  Solution- if unit digit is =2 than 1*2*3*4*5=120. So 120 6 digit numbers can be formed which can divisible by unit digit 2.
  if unit digit is =3 than 1*2*3*4*5=120. So 120 6 digit numbers can be formed which can divisible by unit digit 3.
  if unit digit is =4 than 1*2*3*4*2=48. So 120 6 digit numbers can be formed which can divisible by unit digit 4.(Because last 2 digit must be divisible by 4)
  if unit digit is =5 than 1*2*3*4*5=120. So 120 6 digit numbers can be formed which can divisible by unit digit 5.
  if unit digit is =6 than 1*2*3*4*5=120. So 120 6 digit numbers can be formed which can divisible by unit digit 6.(it is also divisible by 3 because 1+2+3+4+5+6=21 is divisible by 3).
  So total numbers=120+120+48+120+120=528 Ans
• 7 years agoHelpfull: Yes(10) No(6)
• from 1--------120 number
  from 2---------120 number
  from 3----------120 number
  from 5-----------120 number
  from 6------------120 number
  from 4 last two digit should be divisible by 4 so only 2 cases ----24 and ----64
  so total number divisilbe by 4 = 24+24= 48
  so total is -648 answer
• 7 years agoHelpfull: Yes(3) No(6)
• One way to get the answer:
  XXXXX1 is always divisible by 1, so we have 5! numbers.
  XXXXX2 is always divisible by 2, so we have 5! numbers.
  XXXXX3 is always divisible by 3 (sum of digits is always 21), so we have 5! numbers.
  XXXXY4 is divisible by 4 only if Y is 2 or 6, so we have 2 * 4! numbers.
  XXXXX5 is always divisible by 5, so we have 5! numbers.
  XXXXX6 is always divisible by 6 (even number divisible by 3), so we have 5! numbers.
  
  So total number of numbers with required property = 5 * 5! + 2* 4! = 600 + 48 = 648 numbers.
  
  Another way to get the answer:
  Almost all number have the required property except for:
  XXXXY4, where Y is 1 or 3 or 5.
  So 3 * 4! number do not have the required property.
  Number of 6 digits numbers = 6!
  
  So total number of numbers with required property = 6! - 3 * 4! = 720 - 72 = 648 numbers.
• 6 years agoHelpfull: Yes(2) No(1)
• 528 is correct answer
• 7 years agoHelpfull: Yes(1) No(1)
• Without repetition it would be - 648
• 7 years agoHelpfull: Yes(0) No(6)
• unit digit divisibility by 2 and 2 at the unit place = 5! =120 ways eq....1
  unit digit divisibility by 3 and 3 at the unit place = 5! =120 ways eq....2
  unit digit divisibility by 5 and 5 at the unit place = 5! =120 ways eq....3
  
  total 120+120+120=360 ways
  
  unit digit divisibility by 4 and 4 at the unit place covered by eq 1(if no divided by 4 then it must divided
  by 2 also)
  unit digit divisibility by 4 and 4 at the unit place covered by eq 2 and eq 3(divisibility by 6=div by 2 and
  div by 3 )
• 6 years agoHelpfull: Yes(0) No(1)