What is the probability that a 9 digit number formed using 1,2,3,4,5,6,7,8,9 is divisible by 36.(without repetition)

• ans: 2/9
  here the 9digit number formed should be divisible by 36
  36=9*4
  ie the number should be divisible by both 9 nd 4
  1+2+3+...+9 = 45
  so every number formed with these digits can be divisible by 9
  now the last two digits of the number should be divisible by 4
  so last 2 digits can be 12,16,24,28,32,36,48,52,56,64,68,72,76,84,92,98
  total 16 possibilities
  so starting 7 digits can be arranged in 7! ways
  so probability = (16 * 7!)/9!
• 11 years agoHelpfull: Yes(68) No(3)
• @ROJA Probability is asked. Not the number of ways. Probability can never be greater than 1.
• 11 years agoHelpfull: Yes(2) No(2)
• the num which is divisible 36 will always divisible by 9&4 which are co-primes. from the nums 1,2,3,4,5,6,7,8,9 the num divisble by both 9&4 will have last 2 digits as 72or54or36 and the remaining 7 places are arranged in 7! ways.so the ans is 3*7!
• 11 years agoHelpfull: Yes(1) No(5)
• y it can't b 6*6 and divisibility rule for 3 and 2 can be checked and hence 8!*4/9! will b the ans.
• 11 years agoHelpfull: Yes(1) No(3)
• Nice question !!!
• 11 years agoHelpfull: Yes(1) No(2)
• there are 9 places to fill. the last digit can be filled with the digits (2 or 4 or 6 or 8), so there are 4 ways and similarly the second last digit can be filled with (2 or 4 or 6 or 8), i.e that there are four ways.
  and the remaining 7 digits can be filled in 7 ways (7!). therefore, 4*4*7!. in order to find the probability, we divide (4*4*7!) with 9! . THUS, THE ANSWER IS (4*4*7!)/9! i.e., 2/9.
• 6 years agoHelpfull: Yes(0) No(1)