How many 3 digit numbers can be formed using digit 1 to 6,without repetition such that number is divisible by the digit at its unit place ??

• in this case
  _ _ 1 first the last digit is 1 then 5*4=20 cases can be possible
  then
  _ _ 2 last digit is 2 then 5*4 that is 20 cases are possible
  then
  _ _ 3 last digit is three then for the divison of the number by three sum of all the numbers should be multple of three
  then these cases can be possible that are 12,15,21,24,42,51 that are total 6 cases
  then for last digit is four then these cases can be possible that are
  12,16,26,32,36,56,52 and 62 total 8 cases
  for last digit to 5 5*4 cases can be possible
  last digit is 6 then 6 cases that are 12,15,21,24,42 and 51 are possible
  total of 80 cases can be possible
• 9 years agoHelpfull: Yes(1) No(2)
• Number with unit digit 1,2,3,5 and 6 and divisible by the same will be 5*5! i.e 600 ways
  
  now we are left with unit digit 4 and divisible by 4 case so in this case last 2 digits can only be 24 and 64 so 2*4! i.e 48 ways
  
  so total = 600+48 = 648 , it should be i believe
• 6 years agoHelpfull: Yes(0) No(0)