How many 6 digits numbers can be formed by using digits 1 to 6,without repetition such that no is divisible by digit at unit's place
A)402
B)528
C)648
D)720

• C) 648
  All 6 digit numbers ends with:
  '1' are divisible by 1 =120
  '2' are divisible by 2 =120
  '3' are divisible by 3 (as sum of digits=21 which is divisible by 3) =120
  '5' are divisible by 5 =120
  '6' are divisible by 6 (as end digit is 6 so all numbers are divisible by 2 and sum of digits=21 so it is also divisible by 3. Hence, 6)=120
  
  All 6 digit numbers ends with '4' are divisible by 4(number is disivisble by 4 if last two digits are divisible by 4 that is either 24 or 64)=4!*2=48
  
  hence total numbers=120*5+48=648
  
• 10 years agoHelpfull: Yes(32) No(0)
• When last digit 1
  =>all the digits formed will be divisbile by 1
  So no of digits formed is 5!*1
  When last digit 2
  =>all the digits formed will be divisbile by 2 (since all digits will be even cases)
  So no of digits formed is 5!*1
  When last digit 3
  =>all the digits formed will be divisbile by 5(since 1+2+#+4+5+6=21ie divisible by 3)
  So no of digits formed is 5!*1
  When last digit 5
  =>all the digits formed will be divisbile by 5
  So no of digits formed is 5!*1
  similarly
  when last digit 6
  =>all the digits formed will be divisbile by 6
  So no of digits formed is 5!*1
  But when last digit 4
  two cases are formed
  case 1 last 2 digits 24
  no of ways= 4!*1*1
  case 2 last 2 digits 64
  no of ways = 4!*1*1
  so net total no of ways
  last digit{1 ->5!) (2 -> 5!) (3 ->5!) (4 ->4!+4!) (5 -> 5!) (6 -> 5!)
  total = 5! +5! + 5! + 5! +5! + 4! + 4! =648
• 10 years agoHelpfull: Yes(9) No(0)
• lets start with negation of this question. i.e. total no. of such numbers not divisible by unit place.
  
  
  * so at unit place 1,2,3,5,6 cannot be there..
  that is - - - - - 4
  now for that no. not to be divisible by 4 ,other digit next to 4 would must be 1 or 3 or 5 .
  
  - - - - (3 or 4 or 5) 4
  
  now the 4 digits ahead of them would be any no left out of these two digits
  i.e. 4 * 3 * 2 * 1 * 3 * 1 = 72.
  
  total no possible of 6 digits is 6! =720
  
  720 -72 = 648.
  
  Explanation:
  * 1 cannot be at unit place
  2 cannot be at unit place as no. will be even if last no is 2
  3 cannot be at unit place as all digits' sum =21 divisible by 3.
  5 cannot be at unit place as then no .will be divisible by 5.
  6 cannot be at unit place as no. will b of course divisible by 2 if 6 is there & all no. are divisible by 3. as all 6 digit no.s will have sum of their digits= 21 (divisible by 3).
• 10 years agoHelpfull: Yes(2) No(0)
• last digit{1 ->5!) (2 -> 5!) (3 ->5!) (4 ->4!+4!) (5 -> 5!) (6 -> 5!)
  total = 5! +5! + 5! + 5! +5! + 4! + 4! =648
• 10 years agoHelpfull: Yes(1) No(1)