Latest Elitmus Aptitude Question SOLUTION: how many six digit number can be formed using the digits 1 to 6, without repetition, such that the number is divisble by the digit at unit's place

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(#M40008865) ELITMUS Numerical Ability question Keep an EYE

how many six digit number can be formed using the digits 1 to 6, without repetition, such that the number is divisble by the digit at unit's place

Asked In Elitmus amrit bhagat (12 years ago)

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Pargat Pal Singh  (12 years ago)

Total possible nos without condition of divisibility = 6! = 720

only some nos with last digit 4 may not be divisible
which are ending with 14,34,54
no of such nos are 3*4*3*2 = 72

so six digit number can be formed using the digits 1 to 6, without repetition, such that the number is divisble by the digit at unit's place = 720-72= 648

Is this solution Helpfull? Yes (73) | No (16)  

Rakesh Kumar Kohli  (12 years ago)

Ans. 648

As each number will contain all the six digits and the sum of digits is
= 1+2+3+4+5+6 = 21 which is divisible by 3. So each number is divisible by 3. The numbers ending with digit 1 will be divisible 1. The numbers ending with digit 2 will be divisible 2. The numbers ending with digit 3 are divisible 3. The numbers ending with digit 5 will be divisible 5. The numbers ending with digit 6 will be divisible 6. Except the numbers ending with last two digits as 14, 34 and 54 all other numbers ending with 4 are divisible by 4.

The no. of numbers ending with last two digits 14,34 and 54 are
= 3*4*3*2 = 72. (As for a number to be divisible by 4 last two digits must be divisible by 4)
so six digit number that can be formed using the digits 1 to 6, without repetition, such that the number is divisble by the digit at unit's place
= 720-72= 648

Is this solution Helpfull? Yes (53) | No (5)  

Praneeth  (10 years ago)

If units digit has 1 then possible numbers = 120(5!)---------->1
_ _ _ _ _ 1 (remaining 5 digits can be filled in 5! ways
if units digit has 2 then possible numbers = 120-------------->2

if units digit has 3 then possible numbers = 120-------------->3
as the sum of all digits is 21 the number is divisible by 3
if units digit has 4 then possible numbers = 48 (4!*2)
as the last two digits must 24,64 remaining 4 digits can be arranged in 4!*2
if units digit has 5 then possible numbers = 120
if units digit has 6 then possible numbers = 120
total 120+120+120+48+120+120=648

Is this solution Helpfull? Yes (46) | No (2)  

Aravindkumar  (9 years ago)

We will go through this step by step.

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

Is this solution Helpfull? Yes (31) | No   

Ankit Agrahari  (12 years ago)

I have gone with the cases putting 1 2 3 4 5 6 as unit digits
and got the answer as 648(120+120+120+48+120+120)

Is this solution Helpfull? Yes (25) | No (1)  

rahul   (10 years ago)

thanks arjun bhai....great thinking power

Is this solution Helpfull? Yes (14) | No   

gaurav kumar  (11 years ago)

How you are calculating for divisible by 4.
I am not getting 3*4*3*2 = 72.from where 3 *4*3*2 is coming.

Is this solution Helpfull? Yes (10) | No (2)  

SACHIN AGGARWAL  (10 years ago)

6!-4!*3C1=720-72=648

Is this solution Helpfull? Yes (7) | No (1)  

hemanth  (12 years ago)

ans:2468
1 as units place=720
2 as units place=360
3 as units place=720
4 as units place=168
5 as units place=120
6 as units place=360
total six digit numbers=720+360+720+168+120+360=2468

Is this solution Helpfull? Yes (6) | No (66)  

Simran  (10 years ago)

oops i missed something...
If digit place has 1 then possible numbers = 120(5!)
if 2 then possible numbers = 120
if 3 then possible numbers =120
if 4 then possible numbers = 48 (4!*2)
if 5 then possible numbers = 120
if 6 then possible numbers = 120
so total no=648 now it's right!

Is this solution Helpfull? Yes (6) | No (1)  

arjun  (12 years ago)

I THINK 648 NUMBERS ARE POSSIBLE.....

Is this solution Helpfull? Yes (5) | No (17)  

VANI SETHI  (9 years ago)

total possible no =6! ie 720
720/6=120 will b divisible each for 1,2,3,5,6
for 4 divisible its 2/5*120=48( as 14,24,34,44,54 only out of these only 2 numbers are divisible by 4)
so total number =120+120+120+120+120+48=648

Is this solution Helpfull? Yes (4) | No   

MUDIT SHARMA  (10 years ago)

3*4*3*2*1 actually means 3*4!
3 for 14,34,54 which are not divisible by 4.
4! are the ways. So Ans is 720-3*4!=648

Is this solution Helpfull? Yes (3) | No   

Simran  (10 years ago)

If digit place has 1 then possible numbers = 120(5!)
if 2 then possible numbers = 120
if 3 then possible numbers =120
if 4 then possible numbers = 48 (4!*2)
if 5 then possible numbers = 120
if 6 then possible numbers = 20
so total no=648

Is this solution Helpfull? Yes (2) | No (1)  

Anshu kumari  (9 years ago)

when unit place is 1 then 5*4*3*2*1=120
when unit place is 2 then 5*4*3*2*1=120
"""""""""""""""""' 3 then 120
"""""""""""""""""" 4 then 24
"""""""""""""""""" 5 then 120
"""""""""""""""""" 6 then 120
total is 5*120+24=624

Is this solution Helpfull? Yes (2) | No (1)  

Anshu kumari  (9 years ago)

when unit place is 1 then 5*4*3*2*1=120
when unit place is 2 then 5*4*3*2*1=120
"""""""""""""""""' 3 then 120
"""""""""""""""""" 4 then 24*2 bcuz 24 and 64 is divisible by 4
"""""""""""""""""" 5 then 120
"""""""""""""""""" 6 then 120
total is 5*120+24=648

Is this solution Helpfull? Yes (2) | No   

Nirmalrajsanjeev  (8 years ago)

Given number 1,2,3,4,5,6
when unit digit 1 then number ways=5*4*3*2*1=120
number way 1,2,3,5,6=120*5=600
if unit digit 4 then tens place 2,6=4*3*2=24*2=48
total number of ways=600+48=648

Is this solution Helpfull? Yes (2) | No   

rajveer singh  (10 years ago)

it should be 528 because the number that ends with 1,2,3,5 are divisible with 1,2,3,5 respectively but the numbers that ends with 4 and 6 has to have second last digit be 2, so with this we have 48 more numbers which gives total of 528...

Is this solution Helpfull? Yes (1) | No (3)  

sweta kumari  (10 years ago)

u all guys are including 6
bt ony 36 is divisible by 6
wat abt 16,26,46,56?????

Is this solution Helpfull? Yes (1) | No (1)  

Rohit Gangwar  (8 years ago)

digits are divisible by One are=1*5*4*3*2*1=120 (1, , , , ,(2 to 6) )
digits are divisible by Two are=1*4*3*2*1*2=48 (2, , , , ,(4 or 6) )
digits are divisible by Three are=1*5*4*3*2*1=120 (3, , , , ,(1 to 6 expect 3) )
digits are divisible by four are=1*4*3*2*2=48 (4, , , ,(last two digit may be 12 or 16 or 32 or 36 or 52 or 56) )
digits are divisible by Five are=0, because unit digit is 5 and in last should be 5 or 0, we can't repeat
digits are divisible by Six are= divisible by 2 or 3 = 48+120=168
120+48+120+48+0+168=528

Is this solution Helpfull? Yes (1) | No (5)  

sourav pal  (9 years ago)

if the question was as like - "six digit no. can be formed using the digit 0 to 5,without repetition.......... .unit's place" ...then what will be answer a.420 b.426 3.432 d.none ???

Is this solution Helpfull? Yes | No   

M Aslam  (9 years ago)

I have gone with the cases putting 1 2 3 4 5 6 as unit digits
and got the answer as 648(120+120+120+48+120+120)
Read more at http://www.m4maths.com/frequently-asked-placement-questions.php?SOURCE=elitmus&TOPIC=Numerical%20Ability&SUB_TOPIC=Permutation%20and%20Combination#vFws7SICk58fhqWe.99

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