How many 4 digit numbers can be formed using digits 0,1,2,3,4,5, such that number is divisible by 8??

• Applying the divisibility rule of 8 we get
  Case 1: _ 024............(3 ways)
  Case 2: _032............(3 ways)
  Case 3: _104............(3 ways)
  Case 4: _120............(3 ways)
  Case 5: _152............(2 ways)
  Case 6: _240............(3 ways)
  Case 7: _304............(3 ways)
  Case 8: _312............(2 ways)
  Case 9: _320............(3 ways)
  Case 10: _352............(2 ways)
  Case 11: _432............(2 ways)
  Case 12: _504............(3 ways)
  Case 13: _512............(2 ways)
  Case 14: _520............(3 ways)
  So total combinations possible=adding above combinations we get= 3*9 + 2*5= 27+10=37
• 8 years agoHelpfull: Yes(34) No(5)
• No ending with 0 are 4120,3120,5120,1240,3240,5240,1320,4320,5320,1520,3520,4520=12 ways
  no ending with 2 are 1032,4032,5032,3152,4152,4312,5312,1352,4352,1432,5432,3512,4512=13 ways
  no ending with 4 are 1024,3024,5024,2104,3104,5104,1304,2304,5304,1504,2504,3504=12 ways
  so total no would be 12+13+12=37
• 8 years agoHelpfull: Yes(10) No(1)
• A number is divisible by 8 if and only if the last three digit are divisible by 8
  so from the following numbers cases possible
  Case 1: 024 ... remaining digits 1,2,3(3 ways)
  Case 2: 032.... (3 ways)
  Case 3: 104...3 ways
  Case 4: 120...3 ways
  Case 5: 152.. As 0 cant be used so 2 ways
  Case 6: 240.. 3 ways
  Case 7: 304.. 3 ways
  Case 8: 312.. 2 ways
  Case 9: 320.. 3 ways
  Case 10: 352... 2 ways
  Case 11: 432.. 2 ways
  Case 12: 504.. 3 ways
  Case 13: 512... 2 ways
  Case 14: 520.. 3 ways
  Total = 37 Ans
• 8 years agoHelpfull: Yes(6) No(0)
• FIRST taking no's having 0's
  024
  032
  104
  120
  240
  304
  320
  504
  520
  it has 9 possible way now thousand place have 3 ways therefore (9*3)=27
  now taking no with non zeros
  152
  312
  352
  432
  512
  it has 5 possible way now thousand place have 2 ways (since 1st no cant be -ve )therefore (5*2)=10
  so total possible way=27+10=37
  
• 8 years agoHelpfull: Yes(3) No(0)
• write ans will be 60
• 8 years agoHelpfull: Yes(2) No(4)
• the no's can be end with 0,2,4 only
  =>no's end with 0 are 1240,1320,1520,3120,3240,4120,5120,5240,5320 = 9 ways
  =>no's end with 2 are 1032,1352,3152,4032,4132,4312,4512,5032 = 8 ways
  => no's end with 4 are 1024,2104,3104,3024,5024,5104,5304 =7 ways
  total= 9+8+7 =24
• 8 years agoHelpfull: Yes(2) No(3)
• Applying the divisibility rule of 8 we get
  Case 1: _ 024............(3 ways)
  Case 2: _032............(3 ways)
  Case 3: _104............(3 ways)
  Case 4: _120............(3 ways)
  Case 5: _152............(2 ways)
  Case 6: _240............(3 ways)
  Case 7: _304............(3 ways)
  Case 8: _312............(2 ways)
  Case 9: _320............(3 ways)
  Case 10: _352............(2 ways)
  Case 11: _432............(2 ways)
  Case 12: _504............(3 ways)
  Case 13: _512............(2 ways)
  Case 14: _520............(3 ways)
  So total combinations possible=adding above combinations we get= 3*9 + 2*5= 27+10=37
  
• 8 years agoHelpfull: Yes(1) No(0)
• 3*(024)
  3*(032)
  3*(104)
  3*(120)
  2*(152)
  3*(240)
  3*(304)
  2*(312)
  3*(320)
  2*(352)
  2*(432)
  3*(504)
  2*(512)
  3*(520)
  -------------------------
  37 ways
  
  
• 8 years agoHelpfull: Yes(0) No(0)
• Applying the divisibility rule of 8 we get
  Case 1: _ 024
  Case 2: _032
  Case 3: _104
  Case 4: _120
  Case 5: _152
  Case 6: _240
  Case 7: _304
  Case 8: _312
  Case 9: _320
  Case 10: _352
  Case 11: _432
  Case 12: _504
  Case 13: _512
  Case 14: _520
  For all the above cases:Number of numbers formed are: 2!=2
  So total combinations possible= 2 * 14=28
  
• 8 years agoHelpfull: Yes(0) No(2)
• Applying the divisibility rule of 8 we get
  Case 1: _ 024............(3!)=6
  Case 2: _032.............(3!)=6
  Case 3: _104.............(3!)=6
  Case 4: _120.............(3!)=6
  Case 5: _152.............(2!)=2
  Case 6: _240.............(3!)=6
  Case 7: _304.............(3!)=6
  Case 8: _312.............(2!)=2
  Case 9: _320.............(3!)=6
  Case 10: _352.............(2!)=2
  Case 11: _432.............(2!)=2
  Case 12: _504.............(3!)=6
  Case 13: _512.............(2!)=2
  Case 14: _520.............(3!)=6
  So total combinations possible=adding above combinations we get= 6*9 + 2*5 = 54+10=64
  
• 8 years agoHelpfull: Yes(0) No(6)
• can I know why u are not taking 008 & 016?please reply
• 8 years agoHelpfull: Yes(0) No(1)
• 008 is nt taken just becoz no is nt repeated 00 2tyms will come if we take double zero. and 6 no will given why we take 016
• 8 years agoHelpfull: Yes(0) No(0)
• does repeation allow???thn ans wl difrnt
• 8 years agoHelpfull: Yes(0) No(0)
• 38 is correct without repetition...
  Plz let me know whether 135 is the answer if rep is allowed??
  Lile 5*27(cases upto hundreds digit)
• 8 years agoHelpfull: Yes(0) No(1)
• here why dont we take that repeation is allowed????
  can anybody plz explain.....
• 8 years agoHelpfull: Yes(0) No(0)
• in question repetition information isn't given so we assume repetition
  any leave that repetition
  neha gupta and other guys why didn't took 016 so answer is worng
• 7 years agoHelpfull: Yes(0) No(0)