How many three digit numbers can be formed using the digits 0,1,2,3,4,5 only once.( repetition not allowed ) such that the number is divisible by 15?
A) 8 B)14 C )20 D) 25

• ans is 14
  for a number to be divisibl by 15 it should be divisible by both 5 and 3
  divisibility rule of 5: it must end in 5 or 0
  divisibility rule of 3:sum of numbers should be divisible by 3
  numbers are
  345,435,135,315,105,405,450,150,540,240,420,120,210,510
• 6 years agoHelpfull: Yes(34) No(11)
• This question is based on combination ,
  it's solution will be like --
  5C1+5C1+4C1=5+5+4=14
  We have to make 3 digits number so 3 digit number cannot be started from 0 , so we have remaining 5 digits .
  So for 1st digit we have to select 1 digit from any of these 5 digits(1,2,3,4,5) , so it's combination will be - 5C1 .
  Now for 2nd digit - we can consider 0 here but we can't consider the digit which we have used previously so it's combination will be - 5C1 .
  now for 3rd digit - we can consider 0 , but we cannot consider previous 2 digits , so its combination will be - 4C1 .
  Thus , we have - 5C1 + 5C1 + 4C1 = 5+5+4 = 14
• 6 years agoHelpfull: Yes(8) No(13)
• A number is divisible by 15 if it is divisible by 5 so the possible combinations are
  (0,2,3)=there will be 2*2*1=4
  (0,1,4)= there will be 2*2*1=4
  (2,3,5) = there will 3! ways to form the number=3*2*1=6
  (1,4,5) = there will 3! ways to form the number= 3*2*1=6
  so the total number of ways is 4+4+6+6=20
  so option C is correct
• 6 years agoHelpfull: Yes(7) No(30)
• 100 - 200 = 105,120,135,150 total no - 4
  200-300 = 210,240, total no- 2
  300-400= 315,345 total no - 2
  400-500= 45,420,435,450 total no- 4
  500-600= 510,540 total no = 2
  now we can not ahead bcs nxt number will start from 6 and more
  so total number is = 4+2+2+4+3+2 =14
• 6 years agoHelpfull: Yes(7) No(0)
• By using permutation formula:
  n!/(n-r)!
  
  So there is three digit number and 0,1,2,3,4,5 that means total 6
  
  So according to the formula: 6 ! /(6-3)!
  = 120
  Now divide 120/15 = 8
  
  So answer of this question is 8
• 6 years agoHelpfull: Yes(4) No(20)
• UPDATED SOLUTION- BY CORRECTING PREVIOUS DESCRIPTION
  
  Answer is opt B = 14
  # to make 3 digit number 0 can't be placed at hundredth place.
  
  #CASE 1:
  if we place zero at unit place then at tenth and hundredth place we have to put all the combinations of number except zero. i.e.
  _ _ 0 {underscore shows space for number combination}
  hence it can written as 5c2 combination for above blank spaces.
  5c2 = 10
  
  #CASE 2:
  3 digit number is divisible by 15 only if its unit place contains 5 or 0.
  condition for unit digit as 0 is already explains in case 1.
  so case 2 is- Place 5 as unit digit and 0 as tenth digit i.e.
  ==> _ 0 5 {underscore means blank for number combination}
  so, numbers are remains as 1,2,3,4 only that we need to put it on 100th place
  hence,
  4c1 = 4
  # now adding both cases is equals to 10+4= 14
• 6 years agoHelpfull: Yes(3) No(3)
• 14
  as 0 can't be on unit place therefore
  case 1: 0 at the unit place:
  which will give you 5C2
  i.e. 10
  case 2:
  0 at the 100'th place:
  5 will be at unit plce
  then possible combinations will be:
  4c1
  i.e.= 4
  therefore 10+4=14
  option B
• 6 years agoHelpfull: Yes(2) No(0)
• last two digit should be either 0 or 5 .
  so four numbers are left we have to choose by 4c1
  the ans will be 4c1*2=8
• 6 years agoHelpfull: Yes(1) No(0)
• Answer is opt B = 14
  # to make 3 digit number 0 can't be placed at hundredth place.
  
  #CASE 1:
  if we place zero at unit place and at tenth and hundredth place we put all the combinations of number except zero. i.e.
  _ _ 0 {underscore shows space for number combination}
  hence it can written as 5c2 combination for above blank spaces.
  5c2 = 10
  
  #CASE 2:
  3 digit number is divisible by 15 only if its unit place contains 5 or 0.
  condition for unit digit as 0 is already explains in case 1.
  so case 2 is- Place 5 as unit digit and 0 as tenth digit i.e.
  ==> _ 0 5 {underscore means blank for number combination}
  so, for tenth place numbers are remains as 1,2,3,4 only
  hence,
  4c1 = 4
  # now adding both cases is equals to 10+4= 14
• 6 years agoHelpfull: Yes(1) No(0)
• find smallest 3 digit number divisible by 15 is
  105 accept
  120 accept
  135 accept
  150 accept
  find 3 digit smallest number start with 2 and divisible by 15.
  210 accept
  225 wrong
  240 accept
  255 wrong
  find 3 digit smallest number start with 3 and divisible by 15.
  300 wrong
  315 accept
  330 wrong
  345 accept
  find 3 digit smallest number start with 4 and divisible by 15.
  405 accept
  420 accept
  435 accept
  450 accept
  find 3 digit smallest number start with 5 and divisible by 15.
  510 accept
  525 wrong
  540 accept
  555 wrong
  count accepted number
  ans is 14.
• 6 years agoHelpfull: Yes(1) No(1)
• For divisible by 16 it should be divisible by both 5 and 3
  divisibility rule of 5: it must end in 5 or 0
  divisibility rule of 3:sum of numbers should be divisible by 3
  So for 3 Digit number ->
  if 5 at unit place then possible combination 01,04,13
  if 0 at unit place then possible combination 12,15,24,45
  Total 7 and their vice versa 7*2==14 Ans
• 3 years agoHelpfull: Yes(1) No(0)
• the no that is divisible by 15 must be divisible by 3 and 5 both. total no of 3 digit no using these digit when repetition is not allowed=5p1×5p1×4p1=100
  when the last digit is zero of 3 digit no then it must be divisible by 5 but it should also be divisible by 3. then sum of digit should be multiple of 3.
  (1,2,0)= 2p1×1=2
  (1,5,0)=2p1×1=2
  (2,4,0)=2p1×1=2
  (4,5,0)=2P1×1=2
  when 5 is at unit place
  (1,0,5)=1
  (1,3,5)=2p1×1=2
  (4,0,5)=1
  (4,3,5)=2p1×1=2
  total no of 3 digits no that is divisible by 15=8+6=14
• 6 years agoHelpfull: Yes(0) No(1)
• 20
  as unit place can be fill with either 3 or 5
  the rest will be done in 5C2
• 6 years agoHelpfull: Yes(0) No(1)