Latest Elitmus Aptitude Question SOLUTION: How many six digit numbers can be formed using the digits 1 to 6, without repetition such that the number is divisible by the digit at its unit place? Ans is: a) 720 b)

Divisible by 1 and 1 at the unit place:
_ _ _ _ _ 1
This gives us total of 5 x 4 x 3 x 2 x 1 = 120 numbers.

Divisible by 2 and 2 at the unit place:

_ _ _ _ _ 2

This gives us total of 5 x 4 x 3 x 2 x 1 = 120 numbers.

Divisible by 3 and 3 at the unit place:

_ _ _ _ _ 3

Since, any number will have all the digits from 1 to 6 and the sum 1 + 2 + 3 + 4 + 5 + 6 = 21 is divisible by 3

This gives us total of 5 x 4 x 3 x 2 x 1 = 120 numbers.

Divisible by 4 and 4 at the unit place:

_ _ _ _ _ 4

Here, there are two cases
_ _ _ _ 2 4

This gives 4 x 3 x 2 x 1 = 24 numbers.

_ _ _ _ 6 4
This gives 4 x 3 x 2 x 1 = 24 numbers.
This gives us total of 24 + 24 = 48 numbers.

Divisible by 5 and 5 at the unit place:

_ _ _ _ _ 5

This gives us total of 5 x 4 x 3 x 2 x 1 = 120 numbers.

Divisible by 6 and 6 at the unit place:

_ _ _ _ _ 6

As all 6 digit numbers formed with 1 to 6 digits(without repetition) are divisible by 3 and numbers with 6 at the unit place are even.

This gives us total of 5 x 4 x 3 x 2 x 1 = 120 numbers.

None of these cases will have numbers overlapping with each other.

So, Total numbers = 120 + 120 + 120 + 48 + 120 + 120 = 648

The only way this would not be the case, assuming we are working in base ten, is if the tens digit is odd and the ones digit is 4. That gives us 6!-3*4!=648 ways.

This type of question is frequently asked in every elitmus exam.......not exact but similar like that......

(b) 648
when 1 is in unit place ,then the remaining five places can be filled in 5! ways
when 2 is in unit place ,then the remaining five places can be filled in 5! ways
when 3 is in unit place ,then the remaining five places can be filled in 5! ways
when 5 is in unit place ,then the remaining five places can be filled in 5! ways
when 6 is in unit place ,then the remaining five places can be filled in 5! ways
when 4 is in unit place ,then in tens place ,the digit can be either 2 or 6 (in 2! ways)and remaining
four places can be filled in 4! ways . number of ways = 4! * 2!
total number of ways = 5! +5!+5!+5!+5!+(4!*2!)
=648

_ _ _ _ _1=5!=120
_ _ _ _ _2=5!=120
_ _ _ _ _3=5!=120
_ _ _ _ 24=4!=24and_ _ _ _64=4!=24 24+24=48
_ _ _ _ _5=5!=120
_ _ _ _ _6=5!=120
total=648

B
unit place digits :- 1 then total no. of six digits are 120
likewise for 2,3,5,6 we got 120 for each no.
but in case of "4" we have to check unit and tens place for divisibility
then we get 2,4 and 6,4 for last two digits
now we have remaining four position which should be arranged by 4! = 24
2,4 and 6,4 should be in 2 ways
i.e 2*4! =48
total no of six digits should be formed 5*120 +48 =648

Bt for 4 case-3
1-for24
2-for64
3-for32
Hence by all it will b equal to 672

A
4,10
(4)^2+(10)^2=116

if at unit digit place is 1 then 120 numer wiil be formed in which all 120 number will be divisible by 1=>5!=120
if at unit digit place is 2 then 120 numer wiil be formed in which all 120 number will be divisible by 2=>5!=120
if at unit digit place is 3 then 120 numer wiil be formed in which all 120 number will be divisible by 3=>5!=120
if at unit digit place is 4 then 120 numer wiil be formed in which number divisible by 4 is=>4!*2=48
if at unit digit place is 5 then 120 numer wiil be form in which all 120 number will be divisible by 5=>5!=120
if at unit digit place is 6 then 120 numer wiil be form in which all 120 number will be divisible by 6=>5!=120
hence 120*5+48=1248

if last digit is 1 then numbers will be 120
if last digit is 2 then numbers will be 120
if last digit is 3 then numbers will be 120
if last digit is 4 then there will be only two case
if last two digits are 24 and 64 and there will be 24 + 24 total numbers
if last digit is 5 then numbers will be 120
if last digit is 6 then numbers will be 120
hence 120+120+120+120+24+24+120=648

1 present at unit place
number divisible by 1=120
2 present at unit place
number divisible by 2=120
3 present at unit place
number divisible by 3=120
4 present at unit place
number divisible by 4=48
5 present at unit place
number divisible by 5=120
6 present at unit place
numbers divisible by 6=120
hence 120+120+120+48+120+120=648