Latest Elitmus Aptitude Question SOLUTION: How many 6 digits numbers can be formed by using digits 1 to 6,without repetition such that no is divisible by digit at unit's place A)402 B)528 C)648 D)720

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(#M40018518) ELITMUS Numerical Ability question Number System Keep an EYE

How many 6 digits numbers can be formed by using digits 1 to 6,without repetition such that no is divisible by digit at unit's place
A)402
B)528
C)648
D)720

Asked In Elitmus Sumit Rawal (10 years ago)

Unsolved Read Solution (4)

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Arpit Jaiswal  (10 years ago)

C) 648
All 6 digit numbers ends with:
'1' are divisible by 1 =120
'2' are divisible by 2 =120
'3' are divisible by 3 (as sum of digits=21 which is divisible by 3) =120
'5' are divisible by 5 =120
'6' are divisible by 6 (as end digit is 6 so all numbers are divisible by 2 and sum of digits=21 so it is also divisible by 3. Hence, 6)=120

All 6 digit numbers ends with '4' are divisible by 4(number is disivisble by 4 if last two digits are divisible by 4 that is either 24 or 64)=4!*2=48

hence total numbers=120*5+48=648

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Sumit Rawal  (10 years ago)

When last digit 1
=>all the digits formed will be divisbile by 1
So no of digits formed is 5!*1
When last digit 2
=>all the digits formed will be divisbile by 2 (since all digits will be even cases)
So no of digits formed is 5!*1
When last digit 3
=>all the digits formed will be divisbile by 5(since 1+2+#+4+5+6=21ie divisible by 3)
So no of digits formed is 5!*1
When last digit 5
=>all the digits formed will be divisbile by 5
So no of digits formed is 5!*1
similarly
when last digit 6
=>all the digits formed will be divisbile by 6
So no of digits formed is 5!*1
But when last digit 4
two cases are formed
case 1 last 2 digits 24
no of ways= 4!*1*1
case 2 last 2 digits 64
no of ways = 4!*1*1
so net total no of ways
last digit{1 ->5!) (2 -> 5!) (3 ->5!) (4 ->4!+4!) (5 -> 5!) (6 -> 5!)
total = 5! +5! + 5! + 5! +5! + 4! + 4! =648

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surendra bisht  (10 years ago)

lets start with negation of this question. i.e. total no. of such numbers not divisible by unit place.


* so at unit place 1,2,3,5,6 cannot be there..
that is - - - - - 4
now for that no. not to be divisible by 4 ,other digit next to 4 would must be 1 or 3 or 5 .

- - - - (3 or 4 or 5) 4

now the 4 digits ahead of them would be any no left out of these two digits
i.e. 4 * 3 * 2 * 1 * 3 * 1 = 72.

total no possible of 6 digits is 6! =720

720 -72 = 648.

Explanation:
* 1 cannot be at unit place
2 cannot be at unit place as no. will be even if last no is 2
3 cannot be at unit place as all digits' sum =21 divisible by 3.
5 cannot be at unit place as then no .will be divisible by 5.
6 cannot be at unit place as no. will b of course divisible by 2 if 6 is there & all no. are divisible by 3. as all 6 digit no.s will have sum of their digits= 21 (divisible by 3).

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RAHUL BANSAL  (10 years ago)

last digit{1 ->5!) (2 -> 5!) (3 ->5!) (4 ->4!+4!) (5 -> 5!) (6 -> 5!)
total = 5! +5! + 5! + 5! +5! + 4! + 4! =648

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