How many 9 digit numbers are possible by using 1,2,3,4,5 which are divisible by 4, if the repetition is allowed

• last 2 digits can be 12,24,32,44,52(5 ways)
  first 7 places can be filled in 5^7 ways
  Last 2 places can be filled in 5 ways
  So total ways= 5^7 * 5
  =5^8
• 12 years agoHelpfull: Yes(8) No(0)
• Itz 5^8=390625(short-cut method) itz jus simple yaar
  
  First Place Probablity Having A Digit Is 5(like 1,2,3,4,5)
  Since Repetation Allowed Second Place also implies The same so on till 7 location's but the last two digits shud be divisible by 4 and from those 2 locations we can 5 different combination's so the answer for this question is 390625
  
  Actual Formulae :
  5^8=5^7*5
• 12 years agoHelpfull: Yes(5) No(0)
• Division of 4 is sum of last two digit is divisible by 4 so last two digits is 53 35 31 13 22 44 so 6 and remainig 7 digit is 5^7 so totel no is 5^7 * 6=468750
• 12 years agoHelpfull: Yes(4) No(4)
• Itz 5^8=390625
  For the divisibility of 4 we consider last two location as one location.
• 12 years agoHelpfull: Yes(2) No(0)
• Division of 4 is sum of last two digit is divisible by 4 so last two digits is 53 35 31 13 22 44 so 6 and remainig 7 digit is 5^7 so totel no is 5^7 * 6=468750
• 12 years agoHelpfull: Yes(1) No(1)
• Ans: 312500
  
  last two digits can be : 12,24,44,52
  so, 5*5*5*5*5*5*5*4=312500
• 12 years agoHelpfull: Yes(0) No(4)