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

• Required number is 9 digits.Now if the number need to divisible by 4 then the last two digits should be 12,32,44,24 and 52.Now we have 7 places two fill up with 5 numbers.so the combination is (5c1)^7
  and total combination is 5^7*5=5^8...
• 13 years agoHelpfull: Yes(11) No(1)
• ans 5^8
  for number to be divisible by 4 it should end with 12,24,32,44,52
  now for the 1st 7 places we have 5 choices for each so 5*5*5*5*5*5*5 = 5^7
  n for the last 2 places any of the above number (12,24,32,44,52)could come so we again have 5 option so 5^7 *5 = 5^8
• 13 years agoHelpfull: Yes(3) No(1)
• i thnk there are 5 choices for 7 places.
  ans is 5^7*5.
• 13 years agoHelpfull: Yes(2) No(1)
• number divisible by 4 is 12,24,32,44,52;
  we have remaining 7 position .at each position we can put any of 5 digits given .
  so 9 digit number are possible =5^8
  
• 13 years agoHelpfull: Yes(2) No(1)
• for a number to be divisible by 4 the last two digits should be divisible by 4 .
  now out of 1,2,3,4,5.there are only 5 combinations that are divisible by 4
  viz. 12,24,32,44,52.Now for the remaining 7 places we have 7 choices each.
  so .no. 9 digit nos that can be formed=7*7*7*7*7*5
• 13 years agoHelpfull: Yes(0) No(8)
• dont know ...koko
• 13 years agoHelpfull: Yes(0) No(2)