By using 1,2,3,4,5,how many 5 digit no. can be formed which is divisible by 4,repetation of no. is allowed??

• 625 is the correct answer..
  a no divisible by 4 ..if its last two digits are divisible by 4
  so digits divisible by 4 are:- 12,24,32,44,52
  first three digits may be from any of five i.e 5c1*5c1*5c1=125
  So total 5 digit nos=5*5*5*5=625
• 13 years agoHelpfull: Yes(26) No(1)
• 5^(n-1),where n= no of digit
• 13 years agoHelpfull: Yes(15) No(1)
• for first 3 places we have 5*5*5=125 combinations
  next 2 possibilities are 12,24,32,44,52
  so 125*5=625 numbers
• 13 years agoHelpfull: Yes(7) No(2)
• 625
  A no is divisible by 4 if last 2 digits of no are divisible by 4
  So here 12, 24, 32, 44, 52 are last 2 digits(ie 5 nos are possible there)
  Other 3 digits can any of the 5 nos
  So total 5 digit nos=5*5*5*5=625
• 13 years agoHelpfull: Yes(3) No(2)
• 5^3X4= 500
  Because the repetition is allowed First three places starting from MSB can be taken away by each of the five digits which gives
  5X5X5 number of combinations and then in the units place and tens place we can on use the combination (12,32,44,52) because only these combinations can make it divisible by 4.
• 13 years agoHelpfull: Yes(1) No(10)
• the ans is 625 bcz for first 4 digits we have 3 as repetation is allowed
  but for last 2 digits must divide by 4 those are 5(12,24,32,44,52)
  so total is 5*5*5*5-625
• 13 years agoHelpfull: Yes(1) No(0)