How many 4 digits numbers divisible by 5 can be formed with disgits 0,1,2,3,4,5,6 and 6

a) 220
b) 249
c) 432
d) 216

• taking one 6 fixed at first place,
  
  6 _ _ 0 this will give, 6*5=30
  
  6 _ _ 5 this will give, 6*5=30
  
  now confusion of double 6 is gone :)
  
  1 _ _ 0 this can give us 21 (66 is there always)
  2 _ _ 0 this can give us 21
  3 _ _ 0 this can give us 21
  4 _ _ 0 this can give us 21
  5 _ _ 0 this can give us 21
  
  1 _ _ 5 this can give us 21 (66 is there always)
  2 _ _ 5 this can give us 21
  3 _ _ 5 this can give us 21
  4 _ _ 5 this can give us 21
  
  
  so TOTAL = 30+30+(21*5)+(21*4)=249
• 9 years agoHelpfull: Yes(46) No(8)
• If given digits are 0,1,2,3,4,5 and 6.
  Then
  A no. divisible by 5 must end with 0 or 5.
  i.e. ( _ _ _ 0 ) and ( _ _ _ 5 ) can be the no.
  So, ( 6*5*4 + 5*5*4 ) = 220 = Answer.
• 9 years agoHelpfull: Yes(19) No(9)
• 4 digit no which is divisible by 5
  = 6*6*6*2
  =432
  
• 9 years agoHelpfull: Yes(10) No(18)
• repetition is allowed or not ?? else options are wrong..
• 9 years agoHelpfull: Yes(6) No(5)
• @Ram Neek Singh
  1_ _ 0 gives 21 as for avoiding the two 6s confusion lets assume its one 6.
  So the above case gives 5*4 numbers i.e 20 numbers . Now considering 1 more six which causes the only case 1660 , thus made the count 20+1 =21 . Same for every other number at thousandth place . So ans is 249 .
• 9 years agoHelpfull: Yes(6) No(2)
• 249 is the correct ans.
  
• 9 years agoHelpfull: Yes(5) No(4)
• I think option is wrong ,,,since questn given 2 six,,if in questn contain one six then option might match ,,,,but if given 2 six so it cant be answer
  
• 9 years agoHelpfull: Yes(4) No(8)
• we have 4 places [1],[2],[3],[4]
  we know that a no is divisible by 5,if unit place contains either 0 or 5.
  ok now we fix 0 at 4t place then we can fill place 1 with 6 digit,place 2 with 5 digit,place 3 with 4=6*5*4*1=120 ways
  now we fix no 5 at 4t position then we can fill position 1 with 5 way "(insted of 6 cz we can not place 0 in to 1st place else it'll become 3 digit no).so place 2 with 5 way and plae 3 with 4 way=5*5*4=100 ways.
  so total no-120+100=220 Ans
  
  
• 9 years agoHelpfull: Yes(3) No(9)
• 6*5*4*2=240
• 9 years agoHelpfull: Yes(2) No(9)
• Answer is 249. I don't know how but
  Question is right.
  
• 9 years agoHelpfull: Yes(1) No(2)
• A no is divisible by 5 only if it's last place is 0 or 5 so fix those no
  total way for last place is 2
  
  ---0/5.
  
  now for first place except 0 all the nos may come.
  
  total way is 6.
  
  6*6*6*2=432
  is the ans.
• 9 years agoHelpfull: Yes(1) No(5)
• 249 is correct
  assume one 6 then it will give 220
  so the ans must be greater than 220
  now count manually by putting 2 six together it will give 29
  so 220+29=249
• 9 years agoHelpfull: Yes(1) No(1)
• suyash hum aman tera galat hai 249 hi hoga
  
• 9 years agoHelpfull: Yes(1) No(1)
• The number must be divisible by 5 so last digit must be 0/5 ,so for the 4thdigit there is a possibility of 2 chances and remaining 3digits are 6*6*6=216
  
  Then possibilities are 216*2=432.
  
  And=432
• 9 years agoHelpfull: Yes(1) No(3)
• 432 is correct
  
• 9 years agoHelpfull: Yes(1) No(0)
• 4 digit nmbr _ _ _ _ ,,,,divisible by 5 means last digit should be 0 or 5 means 2 ways(_ _ _ 0/5)
  including 0 at first place now total no of way is 6*5*4*2=240
  for 0 at first,,no of way is possible (0 _ _ 0/5) means 5*4=20
  now 240-20=220
  :)
  
• 8 years agoHelpfull: Yes(1) No(0)
• Unit digit can be chose in 2 ways- either 0 or 5.
  
  When unit digit is 5:
  With just a single 6:
  
  Number of numbers possible = 5 * 5 * 4 = 100 (as MSD cannot be a 0)
  
  Now, two 6's can come as 66x 6x6 or x66, and this counts to 5 + 5 + 4 = 14 numbers (for the last case, x can only be from 1-4 and not 0)
  
  So, total numbers ending in 5 = 100 + 14 = 114.
  
  When unit digit is 0:
  With just a single 6:
  
  Number of numbers possible = 6 * 5 * 4 = 120
  
  Now, two 6's can come as 66x 6x6 or x66, and this counts to 5 + 5 + 5 = 15 numbers.
  
  So, total numbers ending in 0 = 120 + 15 = 135.
  
  Thus, total numbers divisible by 6 = 114 + 135 = 249
• 7 years agoHelpfull: Yes(1) No(0)
• 1) place 0 in extreme position filled 3 positrion =5*6*7/2=105
  2) place 5 in extreme position filled remainig 3 posituion= 6*6*5/2=90
  so totAL =195
• 9 years agoHelpfull: Yes(0) No(12)
• Soumil Sharma @Can you explain how you you get 21
  1_ _0 gives 21 how?
  
  @priyanka rastogi tell the steps. Its really confusing question
• 9 years agoHelpfull: Yes(0) No(1)
• correct option is (d)
  bcz here two sixes are given which is same so we divide the no by 2
  6*6*6*2/2=216
• 9 years agoHelpfull: Yes(0) No(2)
• how u ppl are getting 6*6*6 for 1st 3places???
• 9 years agoHelpfull: Yes(0) No(0)