How many 3 digit numbers formed from 2,3,5,6,7 and 9.Which is divisible by 9 and none of the number is repeated.

• For a no. to be divisible by 9, sum of its digits should be divisible by 9...
  3-digit combinations that gives a sum which is divisible by 9 are
  (2,7,9) (3,6,9) (5,6,7)
  So Total no's formed can be 3!+3!+3! = 18
  
  Ans : 18
• 9 years agoHelpfull: Yes(55) No(0)
• (6,7,5)-> 6 ways
  (9,7,2)-> 6 ways
  (9,6,3)-> 6 ways
  ans. 18
• 8 years agoHelpfull: Yes(2) No(0)
• No divisible by 9 sum is divisible by 9
  3 digit combination (279)(369)(567)(225) are possible bz repeation is possible so 3!+3!+3!+3!2!=21
• 9 years agoHelpfull: Yes(0) No(24)
• none of the numbers can be repeated its mentioned so (225) cant b there 18 is the right ans.
• 9 years agoHelpfull: Yes(0) No(0)
• (1st Digit * 2nd Digit * 3rd Digit)
  4 ways * 5 ways * 6 ways = 120.
  
  Solution: 120
• 9 years agoHelpfull: Yes(0) No(8)
• The sum of digits shd b divisible by 9. hence
  (6,7,5)-> 6 ways
  (9,7,2)-> 6 ways
  (9,6,3)-> 6 ways
  with repitation
  (3,3,3) & (6,6,6) &(9,9,9)->3 ways
  answer ->21
• 9 years agoHelpfull: Yes(0) No(4)