How many 4 digit numbers can be formed using 0,1,2,3,4,5 and divisible by 4

• since a number cannot start with 0 so, 1st place of of 4 digit number can be filled in 5 ways i.e., 1,2,3,4,5.
  second place of 4 digit number can be filled in 6 ways i.e., 0,1,2,3,4,5
  since 4 digit numbers should be divisible by 4 so possibility of last 2 places are (04,12 20,24,32,40,52) i.e., 7 possibilities.
  Therefore total number of 4 digut number divisible by 4 are- 5*6*7=210
• 9 years agoHelpfull: Yes(17) No(7)
• since a number cannot start with 0 so, 1st place of of 4 digit number can be filled in 5 ways i.e., 1,2,3,4,5.
  second place of 4 digit number can be filled in 6 ways i.e., 0,1,2,3,4,5
  since 4 digit numbers should be divisible by 4 so possibility of last 2 places are (00,04,12 20,24,32,40,52) i.e., 8 possibilities.
  Therefore total number of 4 digut number divisible by 4 are- 5*6*8=240
  
• 9 years agoHelpfull: Yes(12) No(5)
• HERE WE CAN FORM 4 DIGIT NOS AND THESE ARE DIVISIBLE BY 4 SO LAST 2 DIGITS ARE DIVISIBLE BY 4 SO COMBINATIONS ARE 04,12,20,24,32,40,52(7COMBINATIONS ) SO NO OF NUMBERS ARE 7*4*3(3rd digit is can fill with 4 ways,and 4 th digit is 3 ways )so 7*4*3=84
• 10 years agoHelpfull: Yes(9) No(8)
• @JAGAN is correct but he also counts numbers starting with 0.
  04,12,20,24,32,40,52(7COMBINATIONS ). 04,20,40 = 3*4*3 = 36
  12,24,32,52 = 4*3*3 = 36.
  
  Total = 36+36= 72
• 9 years agoHelpfull: Yes(7) No(1)
• My answer is n=6*6*5*4 = 720 4-DIGIT numbers can be formed. refer the link http://www.algebra.com/algebra/homework/Permutations/Permutations.faq.question.249759.html
• 9 years agoHelpfull: Yes(4) No(8)
• If the repetition of digit is not allowed then ans is 72
  
  
• 9 years agoHelpfull: Yes(3) No(1)
• 240 is the right answer
• 9 years agoHelpfull: Yes(1) No(3)
• total 9 combination s of last two digit 00,20,40,12,32,42,04,24,44 so 9 combination oof last 2 digit
  and in first position zero wont come so 5 possibilty for first place and 2 nd place 6 possibilty
  tottlay === 9 mul 6 mul 5=270
  correct me if am wrong soon bcos i have exam tmrw
• 9 years agoHelpfull: Yes(1) No(3)
• for 04 20 40 last two digit in3 ways ,2 digit in 4ways 1 digit in 3 ways=3x4x3
  for 12 24 32 52 last two digit in 4 ways and if 0 in 2digit then 1 way and 1 digit in 3 ways =4x1x3
  and if 2 digit does not have 0 then 2 digit in 3ways(no 0) and 1 digit(no 0) in 2 ways=4x3x2
  36+12+24=72ans
• 9 years agoHelpfull: Yes(1) No(1)
• 3240
  6c1*6c1*6c1*6c2
• 10 years agoHelpfull: Yes(0) No(3)
• plz help me to solve this question
• 10 years agoHelpfull: Yes(0) No(1)
• pls any one explain it clearly
  
• 10 years agoHelpfull: Yes(0) No(0)
• Ans will be 63 .
  For divisible by 4, last two digit should be divisible by 4 .
  so there two cases first one number two number having with 0 - 04, 20, 40,
  and second case numbers having without 0 - 12 24 32 52,
  so( 3*3*3 ) + (3*3*4) = 63
• 9 years agoHelpfull: Yes(0) No(4)
• rani which is right
• 9 years agoHelpfull: Yes(0) No(0)
• Ans should be 343
  Bcozz 7 combination of last 2 digit no which are divisible by 4 .then 7^n-1=7^4-1=343
• 9 years agoHelpfull: Yes(0) No(4)
• _ _ _ _
  First digit can not be 0 hence First digit can be 5!=120
  Remaining three digits 5*4*3=60
  4 digit numbers can be =120*60=7200
• 9 years agoHelpfull: Yes(0) No(3)
• Solution:
  
  Test of divisibility for 3
  
  The sum of the digits of any number that is divisible by '3' is divisible by 3.
  
  For instance, take the number 54372.
  Sum of its digits is 5+4+3+7+2=21
  As 21 is divisible by '3', 54372 is also divisible by 3.
  
  There are six digits viz., 0,1,2,3,4 and 5. To form 5-digit numbers we need exactly 5 digits. So we should not be using one of the digits.
  
  The sum of all the six digits 0,1,2,3,4 and 5 is 15. We know that any number is divisible by 3 if and only if the sum of its digits is divisible by '3'.
  
  Combining the two criteria that we use only 5 of the 6 digits and pick them in such a way that the sum is divisible by 3, we should not use either '0' or '3' while forming the five digit numbers.
  
  Case 1
  If we do not use '0', then the remaining 5 digits can be arranged in:
  5! ways=120 numbers.
  
  Case 2
  If we do not use '3', then the arrangements should take into account that '0' cannot be the first digit as a 5-digit number will not start with '0'.
  
  The first digit from the left can be any of the 4 digits 1,2,4 or 5
  
  Then the remaining 4 digits including '0' can be arranged in the other 4 places in 4! ways.
  
  So, there will be 4×4! numbers =4×24=96 numbers.
  
  Combining Case 1 and Case 2, there are a total of 120+96= 216, 5 digit numbers divisible by '3' that can be formed using the digits 0 to 5.
  
• 8 years agoHelpfull: Yes(0) No(0)
• Thousand's place cannot be zero-5c1 ways,hundred's place -6c1.
  Now last 2 digits should be multiple of 4 including 00; 04 to 96 and should be less than 100(since 25*4=100) .So,there are 25 possibilities.Therefore total number of ways is 5*6*25=750.
• 8 years agoHelpfull: Yes(0) No(0)
• divisiblity by 4 means last digits must divisible by 4 possibilities (04,12,24,20,32,40,44,52) total 8 possibilities first digit should not be zero then fisrt digit should be 5c1 second digit 6c1
  then ans is 5c1*6c1*8=240
• 8 years agoHelpfull: Yes(0) No(0)
• if repeation is allowed then the ans is=240 otherwise 84
  
• 8 years agoHelpfull: Yes(0) No(0)