How many five digit numbers can be formed by using the digits 0,1,2,3,4,5 such that the number is divisible by 4?

• an s= 144
  divisible by 4 means last two digit should always div by 4.
  unit place=(04,20,40,12,24,32,52)
  _ _ _04
  _ _ _ 20
  _ _ _40
  total no=4x3x2x3=72
  _ _ _12
  _ _ _24
  _ _ _32
  _ _ _52
  total no=3x3x2x4=72
  total n0 = 72+72=144 ans
  
  
• 9 years agoHelpfull: Yes(70) No(15)
• To get a 5 digit no. we must have one of 1,2,3,4,5 in starting---- 5 ways
  remaining 2 places can be filled with 0,1,2,3,4,5 ways-----------6*6 ways
  A no. to be divisible by 4, last 2 digits must be divisible by 4 -----8 ways (04,,12,20,24,32,40,44,52)
  
  
  total: 5*6*6*8=1440
• 9 years agoHelpfull: Yes(14) No(16)
• 144 is correct answer
• 9 years agoHelpfull: Yes(7) No(1)
• 168...w/o repitation
• 9 years agoHelpfull: Yes(7) No(4)
• if no is divisible by 4 the last two digit(unit and tens place) will be 00,04,12,20,24,32,40,44,52 i.e 9 pairs possible and left most place filled in 5 ways and remaining two places filled in 6 ways each.
  so total possible combination=5*6*6*9=1620 (repeatition is allowed).
• 9 years agoHelpfull: Yes(6) No(2)
• 144 IS THE 100% RIGHT ANSWER WHICH SOLVED BY DEEPAK
• 9 years agoHelpfull: Yes(4) No(3)
• any number is divisible by 4 means the last two digits should be divisible by 4
  so...
  possible combinations with the given numbers are 04,12,20,24,32,40,52
  
  case 1: combinations including 0
  04,40,20
  hence 4*3*2 ways are possible for each combination
  hence 4*3*2*3=72 ways
  case 2:
  12,24,32,52
  for these numbers , there are only 3 possible ways to fill first digit place
  hence 3*3*2 ways
  for 4 numbers
  3*3*2*4 =72 ways
  hence ans is 144 ways
  
  
  
• 9 years agoHelpfull: Yes(4) No(1)
• 
  Divisible by 4 means last two digit shouldalways div by 4.
  Last 2 digit set may be=(04,20,40,12,24,32,52)
  case 1=>
  _ _ _04
  _ _ _ 20
  _ _ _40
  total no=4x3x2x3=72
  
  case 2 =>
  _ _ _12
  _ _ _24
  _ _ _32
  _ _ _52
  total no=3x3x2x4=72
  
  total 5 digit no. = 72+72=144 ans
• 8 years agoHelpfull: Yes(4) No(1)
• ans is 600
  
  to be divisble by 4 last two digits should be divisble by 4
  _ _ _ 20 ->4*5*5(bcoz digits can repeat as per the question) =100
  _ _ _40 ->4*5*5=100
  _ _ _12 =100
  _ _ _52 =100
  _ _ _04 =100
  _ _ _24 =100
  
• 9 years agoHelpfull: Yes(1) No(4)
• ans may be 1620.
  1st place 1 to 5 so 5 ways.
  2nd place 0 to 5, so 6 place.
  3rd place 0 to 5, so 6 place.
  4th and 5th place should show that the no is divisible by 4 then 9 ways(00,04,12,20,24,32,40,44,52)
  
  so 5*6*6*9=1620
  
• 9 years agoHelpfull: Yes(1) No(7)
• For the 5-digit integer to be a multiple of 4, the last two digits must form a multiple of 4.
  Options:
  04, 12, 20, 24, 32, 40, 52.
  
  Case 1: Last 2 digits include 0
  Number options for the last 2 digits = 3. (04, 20 or 40.)
  Number of options for the ten-thousands place = 4. (Any of the 4 remaining digits.)
  Number of options for the thousands place = 3. (Any of the 3 remaining digits.)
  Number of options for the hundreds place = 2. (Either of the 2 remaining digits.)
  To combine these options, we multiply:
  3*4*3*2 = 72.
  
  Case 2: Last 2 digits do NOT include 0
  Number options for the last 2 digits = 4. (12, 24, 32, or 52.)
  Number of options for the ten-thousands place = 3. (Any of the remaining 4 digits but 0.)
  Number of options for the thousands place = 3. (Any of the 3 remaining digits.)
  Number of options for the hundreds place = 2. (Either of the 2 remaining digits.)
  To combine these options, we multiply:
  4*3*3*2 = 72.
  
  Total options = 72+72 = 144.
• 7 years agoHelpfull: Yes(1) No(0)
• NOs which ends with 0 24
  nos which ends with 2 24
  nos which ends with 4 24
  total nos 72
• 9 years agoHelpfull: Yes(0) No(17)
• To get a 5 digit no. we must have one of 1,2,3,4,5 in starting---- 5 ways
  A no. to be divisible by 4, last 2 digits must be divisible by 4 -----6 ways (04,,12,20,24,32,40)
  remaining 2 places can be filled in 3C2 ways---------------------------6ways
  
  total: 5*6*6=180.
• 9 years agoHelpfull: Yes(0) No(7)
• 1*3*4*5*1=60 , as at unit place 4 should b fixed and at frst place we cant take zero so nly 1 number. if no. not repeated this will b ansr
• 9 years agoHelpfull: Yes(0) No(9)
• for without repeation
  A no. to be divisible by 4, last 2 digits must be divisible by 4 -----7 ways (04,,12,20,24,32,40,52)
  nos which ends with 04 3*2*1
  nos which ends with 40 3*2*1
  nos which ends with 12 2*2*1
  nos which ends with 24 2*2*1
  nos which ends with 52 2*2*1
  nos which ends with 20 3*2*1
  
  total nos 34
  nos which ends with 04
  
• 9 years agoHelpfull: Yes(0) No(6)
• number of number divisible by 4 ending with 0 = 4*3*2*2(2,4) = 48
  2 = 4*3*2*3(1,3,5) = 72
  4 = 4*3*2*2(2,4) = 48
  total number divisible by 4 = 48+ 72+48 = 168(without repeatation)
• 9 years agoHelpfull: Yes(0) No(3)
• 5*4*3*2*3=360
• 9 years agoHelpfull: Yes(0) No(4)
• 126Ans
  when unit =0
  then ten:2or4
  and rest 3 will
  3*3*2
  total:3*3*2*2:36
  now unit:2then ten:1or3or5
  so rest:3*3*2
  total:3*3*2*3:54
  similarly
  when unit:4
  then total:3*3*2*2:36
  Total sum:36+54+36:126
• 9 years agoHelpfull: Yes(0) No(3)
• To get 5 digit no, we can fill first place with 1,2,3,4,5.....5 ways
  remaining 2 place can be filled with 0,1,2,3,4,5......6 ways
  A no. to be divisible by 4, last 2 digits must be divisible by 4....9 ways (00,04,12,20,24,32,40,44,52)
  Total no=9(5x6x6)=1620 ans
• 8 years agoHelpfull: Yes(0) No(1)