How many four digit numbers can be formed using the digits 1, 2,3,4 ,5 ( but with repetition) that are divisible
by 4?

• a no. is divisible by 4 , if no. formed by last 2 digits is divisible by 4.
  there are 5 combinations- 12,24,32,44,52
  1st 2 digits can be filled in 5*5 ways
  total 4 digits no. = 5*5*5=125
  
• 9 years agoHelpfull: Yes(7) No(0)
• For divisibility by 4 last two digits must be divisible by 4 here five such combinations are there 32,44,52,12,24 and as thus first two digits for every combination can be selected in 5*5 ways so it total 5*5+5*5+5*5+5*5+5*5=125
  answer is 125
• 9 years agoHelpfull: Yes(1) No(0)
• last two digit will be divisible by 4
  so last two digit divisible by 4=12,32,52,44,24
  and else may be any digit
  so through permutation
  total no. of digit=5*5*5*(no. of last two digit divisible by 4)
  5*5*5*5
  625
  total no.=625
• 9 years agoHelpfull: Yes(1) No(1)
• last two nos. must be divisible by 4, the nos. are 32,24,12,52,44=5 numbers
  any 5 no.for hundred place(1,2,3,4,5)=5 numbers
  any 5 nos. for thousand place(1,2,3,4,5)=5 numbers
  total 4 digit no is 5*5*5=125
  
• 9 years agoHelpfull: Yes(1) No(0)
• As there repetition is allowed so the
  5_5_12
  _24
  _32
  _52
  _44
  so 5*5*5=125
• 9 years agoHelpfull: Yes(0) No(0)
• 5*5*5*2=250
• 9 years agoHelpfull: Yes(0) No(0)
• if num is divisible by 4 , if no. formed by last 2 digits is divisible by 4.
  there are 5 combinations- 12,24,32,44,52
  1st 2 digits can be filled in 5*5 ways
  total 4 digits no. = 5*5*5=125
• 9 years agoHelpfull: Yes(0) No(0)