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

• 5^8
  last two is divisable by 4 is condition.
  the possibility is 5(12,24,44,32,52)
  5*5*5*5*5*5*5*5
• 13 years agoHelpfull: Yes(62) No(26)
• 5^N-1 is da formula
• 13 years agoHelpfull: Yes(48) No(10)
• In order to be divisible by 4, the two least significant digits must be
  12, 24, 32, 44, 52 (a total of 5 choices)
  The other 7 digits can be any of the five allowed numbers..
  The total number of possible number is
  (5^7)*(5) = 5^8 = 390625..
• 12 years agoHelpfull: Yes(32) No(6)
• 5^8
  will be the ans
• 13 years agoHelpfull: Yes(14) No(10)
• last 2 digit should be divisible by 4, for divisibility by 4
  so last two can be 12 24 32 44 52
  now out of left seven places each can be filled in 5 way
  
  therefore ans=5*5*5*5*5*5*5*4
• 13 years agoHelpfull: Yes(9) No(37)
• If repetition is allowed, first 7 places can be filled in 5^7 ways and last 2 places can be 12, 24, 32, 52, 44 i. e. 5 ways therefore answer is 5^8
• 10 years agoHelpfull: Yes(3) No(3)
• please explain it how the ans of it can be find out.
• 12 years agoHelpfull: Yes(1) No(5)
• If A number has to be divisible by 4, the last two digits must be divisible by 4. So possibilities are, 12, 24, 32, 44, 52. And the of the remaining 7 places, each place got filled by any of the five digits. So these 7 places got filled by 5 x 5 x .....(7 times) = 5^7 ways. So total ways are 5 x 5^7 = 5^8
• 9 years agoHelpfull: Yes(1) No(0)
• The number which is divisible by 4 has 0,2,4,6,8 in its unit place.
  so the unit place can be filled by 2 different ways , the first digit in 4 ways and rest 8 digits in 5 ways .
  Total number of ways =4*5^8*2=3125000
• 8 years agoHelpfull: Yes(0) No(0)
• 5 ^ 8 * 2 =781250
• 8 years agoHelpfull: Yes(0) No(0)
• last two digits must be divisible by 4. So possibilities are, 12, 24, 32, 44, 52.
  7 places, are left to fill by any of the five digits in this way we get
  5*5*5*5*5*5*5*5=5^8
• 7 years agoHelpfull: Yes(0) No(0)