how many 6 digits numbers can be formed using 0 to 5 without repition so that the number will be divisible by its unit digit

• the number ends with 1,2,3,4,5
  so how many number can be made using given six digits whose last value is 1.
  4*4*3*2*1 1(last digit is fix)=96
  simillarly for 2,3,4,5 we will get total 96.
  so total nuber that we can form is 96**5=480;
  96*5=480;
  in every case it is divisible by last unit place's num except 4;
  in 4 some of the number that is not divisible are given bellow
  the number which ends with 14,34,54
  so for this
  3*3*2*1 (14) is fixed
  3*3*2*1 (34) is fixed
  3*3*2*1 (54) is fixed
  so total is 3*(3*3*2*1)=54
  
  so final solution is:- 480-54=426
  
• 11 years agoHelpfull: Yes(21) No(6)
• Ans:420 6-digit numbers.
  
  0- as units digit: No numbers can be formed as the units digit should divide the no.
  put,
  1- in units place: 5 places are remaining _ _ _ _ _ 1 .First digit shouldn't be a '0' so there are (5!-4!)*1 = 96 ways(numbers).
  2- in units place: 5 places are remaining _ _ _ _ _ 2 .First digit shouldn't be a '0' so there are (5!-4!)*1 = 96 ways(numbers).
  3- in units place: 5 places are remaining _ _ _ _ _ 3 .First digit shouldn't be a '0' and the number must be divisible by 3. Sum of all digits 1+2+3+4+5=15. Since its a non-repeating number , it will be divisible by 3. So,there are (5!-4!)*1 = 96 ways(numbers).
  4- in units place: 5 places are remaining _ _ _ _ _ 4 .First digit shouldn't be a '0' and the number must be divisible by 4. So, the ten's digit must be either 2 or 4 i.e., 2 ways to fill ten's place. Now 4 places are empty keeping aside units& tens. These three empty places can be arranged in (4!-3!)*1 . Total ways =(4!-3!)*2*1 = 36 ways(numbers).
  5- in units place: 5 places are remaining _ _ _ _ _ 5 .First digit shouldn't be a '0' and here any digits from 0 to 4 excluding '0' from the 6th place, there are (5!-4!)*1 = 96 ways(numbers).
  
  Hence Total numbers would be : 96 + 96 + 96 +36 +96 = 420.
  
• 11 years agoHelpfull: Yes(9) No(5)
• unit digit=1 _ _ _ _ _ 1=4*4!=96
  unit digit=2 _ _ _ _ _ 2=4*4!=96
  unit digit=3 _ _ _ _ _ 3=4*4!=96
  unit digit=4(last 2 digit must be divisible by 4) {last 2 digit 12 20 24 32 40 52} _ _ _ _ 12/24/32/52 +_ _ _ _ 20/40= 3*3!*4 + 4!*2=120
  unit digit=5 _ _ _ _ _ 5/0= 4*4!+5!=120
  
  total 96*3+120*2=528
• 11 years agoHelpfull: Yes(2) No(5)
• 78 nos can be formed,,,
  to be divisible by 4 last 2 digit shoul be 2,4 rest digits can be arranged in 3*2*1*1*1 i.e 6 nos of this form
  to be divisilble by 2 last digit should be 2 rest digits can be arranged in 4*3*2*1*1 i.e again 24 no of such form
  to be divisilble by 5 last digit should be 5 rest digits can be arranged in 4*3*2*1*1 i.e again 24 no of such form
  to be divisible by 3 sum of digits should be divisible by 3 sum is 15 i.e 24 no of digits are possible cos 5th place can be arranged in 4 ways cos unit digit is fixed as 3 4th place in 3 ways 3rd place in 2 ways 2nd place in 1 ways and unit place is fixed with 3 so 4*3*2*1*1=24 , 6+24+24+24( rest divisible no)=78 nos can be formed
• 11 years agoHelpfull: Yes(1) No(7)
• /*correction in last ans*/
  unit digit=1 _ _ _ _ _ 1=4*4!=96
  unit digit=2 _ _ _ _ _ 2=4*4!=96
  unit digit=3 _ _ _ _ _ 3=4*4!=96
  unit digit=4(last 2 digit must be divisible by 4) {last 2 digit 12 20 24 32 40 52} _ _ _ _ 12/24/32/52 +_ _ _ _ 20/40= 3*3!*4 + 4!*2=120
  unit digit=5 _ _ _ _ _ 5/0= 4*4!+5!=216 //calculation mistake here
  
  total 96*3+120+216=624
• 11 years agoHelpfull: Yes(1) No(4)
• num end with 1=96
  with 2=96
  with 3=96
  with 5=96
  with 0=96(bcoz all nos ends with these digits r divisible by them)
  nos ends with 4=24(ends with 04)+18(ends with 24)=42
  so total nos=522
  
• 11 years agoHelpfull: Yes(1) No(0)
• 27 nos can be formed,,,
  to be divisible by 4 last 2 digit shoul be 2,4 i.e 1 nos of this form
  to be divisilble by 2 last digit should be 2 i.e again 1 no of such form
  to be divisible by 3 sum of digits should be divisible by 3 sum is 15 i.e 24 no of digits are possible cos 5th place can be arranged in 4 ways cos unit digit is fixed as 3 4th place in 3 ways 3rd place in 2 ways 2nd place in 1 ways and unit place is fixed with 3 so 4*3*2*1*1=24 , 24+3( rest divisible no)=27 nos can be formed
• 11 years agoHelpfull: Yes(0) No(7)
• ans is - 984
  to be divisible by 1 , 2,5 =4*4*3*2*1*1=96 =96*3=288
  to be divisible by 3=5*5*4*3*2*1=600
  to be divisible by 4=4*3*2*1*4(last2)=96
• 11 years agoHelpfull: Yes(0) No(4)
• To form any 6 digit num using given 0 must not be placed in first place as it becomes a 5 digit num now place 0 in last so places available are five first place can be filled in 5 ways is repitation is not allowed second place will be filled in 4 ways followed by 3 2 1 thus 5*4*3*2*1=120 all the num are divisible by zero if 1 is placed in lst place again we get 120 num's lly for 2 and 3 in all this cases number is divisible by last digit in case 4 is in last place as num must be divided by 4 we can place only 0 or 2 in tens place thus we get 96 num divisible by 4 if 5 is placed we get again 120 num's total=(120)4+96=576 numbers
• 11 years agoHelpfull: Yes(0) No(2)
• let unit digit is 1 then total nos = 5!-4!;
  when unit digit is 2 then total nos= 5!-4!;
  when unit digit is 3 then total nos are 5!-4!;
  when unit digit is 4 then ten digit must b either 0 or 2 so total nos are 4!+4!-3!;
  when unit digit is 5 then total nos are 5!-4!
  total nos are 9234.
• 11 years agoHelpfull: Yes(0) No(1)
• numbers divisible by 1,2,3,5 are 96*4=384
  and by unit digit as 4 are 18
  so total numbers are 384+18= 402
  is this a correct answer?
  
  
  
  
  
  
  
  
  
  
  
• 11 years agoHelpfull: Yes(0) No(1)