Latest Elitmus Aptitude Question SOLUTION: How many 6 digit numbers can formed by 0-5 digits divisible by digit at unit place? options1-420 2-426 3-432,none

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(#M40033908) ELITMUS Numerical Ability question permutation Keep an EYE

How many 6 digit numbers can formed by 0-5 digits divisible by digit at unit place? options1-420 2-426 3-432,none

Asked In Elitmus shuvaranjan dash (10 years ago)

Unsolved Read Solution (3)

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Read Solution (3) :

Pu  (10 years ago)

1 at unit place=>4*4*3*2*1=96
2 at unit place=>4*4*3*2*1=96
3 at unit place=>4*4*3*2*1=96
5 at unit place=>4*4*3*2*1=96 but for 4 the last 2 digit must be divisible by 4 so, we have 2 options 0 and 2 at second place and 4 at unit place. now,
when 4 at unit place and 2 at second place=>3*3*2*1*1=18
and 4 at unit place and 0 at second place=>4*3*2*1*1=24
Now, 96*4+18+24=426

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Prateek Shinde  (10 years ago)

_ _ _ _ _ 1 => one at unit place, all 6digit number are divisible by one,
but 1 should be at once place and we can manipulate rest 5 digits in 5! ways
therefore 5!=120
_ _ _ _ _ 2 => 2 at units place, all number ending with 2 are divisible by 2,
therefore 5!=120
_ _ _ _ _ 3=> 1+2+4+5+6+3=21 divisible by 3
therefore 5!=120
_ _ _ _ (2/6) 4 => 4 at unit place and either 2 or 6 in tens place make the whole number divisible by 4, rest 4 digits in 4! ways
therefore 4!*2=48
_ _ _ _ _ 5 => 5!= 120
_ _ _ _ _ 6 => 5!= 120

total comes up to 120+120+120+120+120+48=648

Is this solution Helpfull? Yes | No (11)  

Manisha Choudhary  (7 years ago)

Ans is 426.
focus on zero(As leftmost digit cannot be zero)
a)- - - - - 1
4 4 3 2 1 1 -> divisibility by 1
total = 96
b)- - - - - 2
4 4 3 2 1 2 -> divisibility by 2
Total = 96
c)- - - - - 3
4 4 3 2 1 3 -> divisibility by 3
Total 96
d)- - - - - 4
Here, take 2 cases for divisibilty by 4.
- - - - 2 4
3 3 2 1 2 4
Total 18
Another case as,
- - - - 2 0
4 3 2 1 2 0
Total = 24
e) - - - - - 5
4 4 3 2 1 5 -> divisibilty by 5
Total = 96

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