How many 3 digit odd number can be formed which are divisible by 5 using the numbers from 0 to 9(without repetition).

• If the no is divisible by 5 then the last digit must be 0 or 5.
  But since no has to be odd so last digit will be only 5(fixed).
  First digit can be 1,2,3,4,6,7,8,9=8digits
  Second digit can be 0 also so there can be 8digits.
  ans=8*8*1=64
• 7 years agoHelpfull: Yes(16) No(3)
• ans 56
  8n2=56
• 10 years agoHelpfull: Yes(6) No(0)
• _ _ _
  (1)(2) (3)
  (3)--> 2 ways i.e 0,52
  (2)--> 9
  (1)-->8
  therefore 2*9*8 ways= 144
• 10 years agoHelpfull: Yes(3) No(4)
• ans: 12
  three digit odd no from 0-9 are( without repetition)
  135,175,195
  315,375,395
  715,735,795
  915,935,975 so total no are 12
• 11 years agoHelpfull: Yes(0) No(12)
• last two digits must be 0, 5 so that it can be divisible by 5. So those two numbers can be arranged in 2p1 ways... i.e. 2
  Rest two digits can be arranged in 10p2 ways... if rep is allowed.
  Sp total combo: 10p2* 2p1=180 ans
• 10 years agoHelpfull: Yes(0) No(5)
• ans 100
  (5,15,25,35,45,55,65,75,85,95)= 10*10=100 numbers
• 7 years agoHelpfull: Yes(0) No(3)
• 0,1,2,3,4,5,6,7,8,9=total 10 digits
  
  _,_,_
  3rd place an be filled by=1/3/5/7/9 so total=5 ways
  1st place 0 cant be there coz it will be 2 digit no.
  so excluding last digit we have (10-1)=9 digit remains
  so 0 exclude remaining=8 digits
  so 1st place can takes 8 digit and 2nd place can take 7 digit(in this case it takes 8 coz 0 in the middle is allowed)
  ans =8*8*5=320
  Enjoy
• 7 years agoHelpfull: Yes(0) No(3)
• There are three places _ _ _
  Now the first two places can be anything so 9p2
  Now the last one can be only 5,0 because divisble by 5 so 2c1
  So ans is 5c2*9p2
• 5 years agoHelpfull: Yes(0) No(0)