how many 6 digit no. can be formed using digits 0 to 5,without repetition such that number is divisible by digit at its unit place?
a)420
b)426
c)432
d)none of above

• b) 426
  when unit place is 1 then numbers which are divisible by 1 = 96 (4*4*3*2*1*1 ways)
  when unit place is 2 then numbers which are divisible by 2 = 96 (4*4*3*2*1*1 ways)
  when unit place is 3 then numbers which are divisible by 3 = 96 (4*4*3*2*1*1 ways)
  when unit place is 3 then numbers which are divisible by 4 = 42 (3*3*2*1*1*1 ways when last two digits are 24 and 4*3*2*1*1*1 ways when last two digits are 04)
  when unit place is 3 then numbers which are divisible by 5 = 96 (4*4*3*2*1*1 ways)
  total=426
  
• 9 years agoHelpfull: Yes(48) No(6)
• b) 426
  when unit place is 1 then numbers which are divisible by 1 = 96 (4*4*3*2*1*1 ways)
  when unit place is 2 then numbers which are divisible by 2 = 96 (4*4*3*2*1*1 ways)
  when unit place is 3 then numbers which are divisible by 3 = 96 (4*4*3*2*1*1 ways)
  
  when unit place is 4 and ten place is 2 means (24) then numbers which are divisible by 4 = 18 (3*3*2*1*1*1 ways )
  when unit place is 4 and ten place is 0 means (04)then number which are divisible by 4 also= 24 (4*3*2*1*1*1 ways )
  
  
  when unit place is 3 then numbers which are divisible by 5 = 96 (4*4*3*2*1*1 ways)
  total=96+96+96+18+24+96=426
  ans
  
  same as Vikas Jha solution but more clear
• 9 years agoHelpfull: Yes(14) No(2)
• Can u pls tell me why r u not considering 0 at unit place.???
• 9 years agoHelpfull: Yes(6) No(0)
• (b) 426
  we have given digits 0,1,2,3,4,5
  we can't use 0 at lak place(from left first digit)
  when unit place is 1 then numbers which are divisible by 1 = 96 ways (4*4*3*2*1*1)
  when unit place is 2 then numbers which are divisible by 2 = 96 ways (4*4*3*2*1*1)
  when unit place is 3 then numbers which are divisible by 3 = 96 ways (4*4*3*2*1*1)
  when unit place is 5 then numbers which are divisible by 5 = 96 ways (4*4*3*2*1*1)
  when unit place is 4 and tens place 0 then numbers which are divisible by 4= 18 ways (3*3*2*1*1*1)
  when unit place is 3 then numbers which are divisible by 3 = 24 ways (4*3*2*1*1*1)
  total number which unit digit is 4=24+18=42
  total number is =4*96+42=384+42=426
  
  
  
  
• 9 years agoHelpfull: Yes(5) No(0)
• b) 426
  when unit place is 1 then numbers which are divisible by 1 = 96 (4*4*3*2*1*1 ways)
  when unit place is 2 then numbers which are divisible by 2 = 96 (4*4*3*2*1*1 ways)
  when unit place is 3 then numbers which are divisible by 3 = 96 (4*4*3*2*1*1 ways)
  
  when unit place is 4 and ten place is 2 means (24) then numbers which are divisible by 4 = 18 (3*3*2*1*1*1 ways )
  when unit place is 4 and ten place is 0 means (04)then number which are divisible by 4 also= 24 (4*3*2*1*1*1 ways )
  
  
  when unit place is (5) then numbers which are divisible by 5 = 96 (4*4*3*2*1*1 ways)
  total=96+96+96+18+24+96=426
  ans
  
  IN PREVIOUS SOLUTION I HAD WRITTEN 3 IN PLACE OF 5
• 9 years agoHelpfull: Yes(5) No(0)
• Can anyone tell me hw a no end with 3 divisble by 5
• 9 years agoHelpfull: Yes(4) No(0)
• I think it would be 426.The case with zero at unit digit will not be considered.
• 9 years agoHelpfull: Yes(3) No(0)
• ans is 438
  when unit place is 1 then numbers which are divisible by 1 = 96
  when unit place is 2 then numbers which are divisible by 2 = 96
  when unit place is 3 then numbers which are divisible by 3 = 96
  when unit place is 4 then numbers which are divisible by 4 = 54
  when unit place is 5 then numbers which are divisible by 5 = 96
  total =438 ans
  
• 9 years agoHelpfull: Yes(2) No(11)
• its 432....
• 9 years agoHelpfull: Yes(2) No(2)
• 426 is correct answer
• 9 years agoHelpfull: Yes(1) No(0)
• questions are same but digits are 1 to 6 given...solve plzz...
• 9 years agoHelpfull: Yes(1) No(0)
• it should be 420
  
  when unit place is 1 then numbers which are divisible by 1 = 96
  when unit place is 2 then numbers which are divisible by 2 = 96
  when unit place is 3 then numbers which are divisible by 3 = 96
  when unit place is 4 then numbers which are divisible by 4 = 36
  when unit place is 5 then numbers which are divisible by 5 = 96
  total=420
• 9 years agoHelpfull: Yes(0) No(6)
• b) 426
  units digit 1 = 4(4!) = 96
  units digit 2 = 4(4!) = 96
  units digit 3 = 4(4!) = 96
  units digit 4 = 3(3!)+3(3!)=36 {divisible by 4( last two numbers should be divisible by 4 ie. 04 & 24 only possible ways) }
  units digit 5 = 4(4!) = 96
  4(96)+36 = 420
• 9 years agoHelpfull: Yes(0) No(7)
• sorry for the previous solution it was wrong assumption the answer is
  d)none of the above
  units digit 1 = 4(4!) = 96
  units digit 2 = 4(4!) = 96
  units digit 3 = 4(4!) = 96
  units digit 4 = 4!+7[3(3!)]=150 {divisible by 4( last two numbers should be divisible by 4 ie. 04,12,20, 24,32,40,52 are possible ways) }
  units digit 5 = 4(4!) = 96
  4(96)+150=534
• 9 years agoHelpfull: Yes(0) No(7)
• See @Vamsi when you consider cases like 12,32,52 and so on for last two digit..then you have to check the divisibility by 2(no which is at unit place) .You only have to consider two cases 04&24 for the last two digit,Such that number is divisible by 4.
• 9 years agoHelpfull: Yes(0) No(2)
• A number is divisible by 3 iff the sum of the digits is divisible by 3 .
  How come divisibility by 3 goes as 4*4*3*2*1*1 ? @vikas jha?
• 9 years agoHelpfull: Yes(0) No(0)
• 5*4*3*2*1*2=240 ans(d)
• 9 years agoHelpfull: Yes(0) No(1)
• questions are same but digits are 1 to 6 given...solve plzz..
  
  Hello mamta Kumari....If u want to solve digits 1 to 6 then ans is 648.
  
  
  120+120+120+48+120+120=648.
• 7 years agoHelpfull: Yes(0) No(0)
• how it is 4*4*3*2*1*1
• 5 years agoHelpfull: Yes(0) No(0)