Latest Elitmus Aptitude Question SOLUTION: How many six digit numbers can be formed using digit 1 to 6, without repition such that the number is divisible by the digit in it's unit place a.402 b.528 c.648 d.720

First we fix 1 at unit place of a six digit number.
Since any number can be divided by 1 .
So it does not matter what comes at other five places .
_ _ _ _ _ 1
So, first position can be filled in 5 ways as any number from 2 to 6 can be placed in that position . Similarly 2nd position can be filled with 4 ways .
So , by fundamental principle of counting ,
We get,

5x4x3x2x1x1=120 or 5! x 1=120

Now for 2 at units place ,

Any even number is divided by 2 ,
Hence it is again similar to first case,
5! × 1 =120

Now for 3 at units place ,

For any number to be divisible by 3 , we need to make sure that sum of the digits is divisible by 3.
Now , 1+ 2 + 3 +4 + 5 +6 = 21
Since 21 is divisible by 3 , so any number formed by number 1 to 6 is divisible by 3 .
Hence , 5! × 1 =120.

Similarly for 5 , 5! ×1 = 120

Now for 6 at unit place ,
We know that any number which is divisible by 2 and 3 is also divisible by 6 .

Hence , 5! ×1 =120

Now for 4 at unit place.
If there is even number at tens place with 4 at units place , then the number is divided by 4 else no.

So we have only two even number 2 and 6 which we can put at tens place .
So now first fixing 2 at tens place and 4 at units place .
_ _ _ _ 2 4

We get by fundamental principle of counting ,
4 × 3 × 2 × 1 × 1 ×1
Or
4! × 1 ×1 = 24

Similarly for 6 at tens place and 4 at units place .
We get 4! × 1 × 1 = 24

Now let us add all the values we got ,


120 +120 +120 + 120 +120 +24 +24 = 648

Hence answer is (c)

1 units place divisable by 1 so 5!=120
2 units place is divisable by 2 so 5!=120
3 units place is divisable by 3 so 5!=120
4 units place is divisable by either 2 or 6=(2*4!)=48
5 units place is divisable by 5 so 5!=120
6 units place is divisable by (3*2) so 5!=120
so answer is 120+120+120+48+120+120=648

option c ..
if we take 1 at unti place then whole number is divisible by 1---so total combination ---5!=120
if we take 2 at unti place then whole number is divisible by 2---so total combination ---5!=120
if we take 5 at unti place then whole number is divisible by 5---so total combination ---5!=120
if we take 6 at unti place then whole number is divisible by 6(bcz a num is divisible by 6 iff it iss divisible by 2 and 3..and if we take 6 at unit digit then it is divisible by 2 and rest of digit is 1+2+3+4+5=15 divisible by 3.(we can take it in any order))---so total combination ---5!=120
if we take 3 at unti place then whole number is divisible by 3(if unit digit is either 3 or 6...and 6 we already considered so only option at unti digit is 3..)---so total combination ---5!=120
if we take 4 at unti place then whole number is divisible by 4(then last 2 digit must be divisible by 4 ..and here at last 2 digit only 2 option 24 and 64)---so total combination ---4!*2=48
.......................................................................
total 648

ans is 528
1 present at unit place =120
2 present at unit place=120
3 presenr at unit place=120
4 present at unit place=24+24=48
5 present at unit place=120
there are no such numbers which is divisible by 6
ans =528

coz 52 and 32 does not ends with 4

First we fix 1 at unit place of a six digit number.
_ _ _ _ _ 1
_ _ _ _ _ 1
_ _ _ _ _ 1
_ _ _ _ _ 1
_ _ _ _ _ 1

when we put 1 at unit place- 5!
when 2 -5!
when 3 -5!
when 4 -4!*2
when 5 -5!
when 6 -5!
so total = 5*5!+4!*2=648

guys all of you are taking 24 and 64 but 52 and 32 are also there what about them