Latest TCS Aptitude Question SOLUTION: 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

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.

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

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

/*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

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

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

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

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

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.

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?