Latest TCS Aptitude Question SOLUTION: By using 1, 2, 3, 4, 5 how many 5 digit no. can be formed which is divisible by 4 repetition of no. is allowed?

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By using 1, 2, 3, 4, 5 how many 5 digit no. can be formed which is divisible by 4 repetition of no. is allowed?

Asked In TCS venkatesh (12 years ago)

Solved Pargat Pal Singh Read Solution (9)

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Read Solution (9) :

Pargat Pal Singh  (12 years ago)

Last two digits can be 12,24,32,44,52.... 5 combinations

First 3 digits can have any of 5 values i.e. 1,2,3,4,5.

so total possible 5 digit nos are 5*5^3 = 5*5*5*5 = 625

Is this solution Helpfull? Yes (20) | No (1)  

prasad  (12 years ago)

we have to get a 5 digit number that is divisible by 4
-----
in this last two digits can be 12,24,32,44,54
so last two digits can be filled in 5 ways
and remaining three
1 st digit can be from 1 to 5 so it can be filled in 5ways
similarly 2nd digit and 3rd digit also 5 ways
so answer is 5*5*5*5=625

Is this solution Helpfull? Yes (9) | No (2)  

Rakesh Kumar Kohli  (12 years ago)

I have earlier solved puzzle considering repetition not allowed.

If repitition is allowed then the last two digit of 5 digit no. divisible by 4 are 24, 32, 12 or 52 and 44.
So number 5 digits no. divisible by 4 can be formed is 5*5*5*5 = 625

Is this solution Helpfull? Yes (4) | No (1)  

prasad  (12 years ago)

we have to get a 5 digit number that is divisible by 4
-----
in this last two digits can be 12,24,32,44,52
so last two digits can be filled in 5 ways
and remaining three
1 st digit can be from 1 to 5 so it can be filled in 5ways
similarly 2nd digit and 3rd digit also 5 ways
so answer is 5*5*5*5=625

Is this solution Helpfull? Yes (4) | No   

Rakesh Kumar Kohli  (12 years ago)

Ans. 12

For divisible by 4 the last two digits of the number should be divisible by 4.
So the numbers must have last two digits as 24 or 32.
If last two digits are 24 then digits 1,3,5 are to be arranged arranged in 6 ways to make numbers as 13524,15324,31524,35124,51324,53124
If last two digits are 32 then digits 1,4,5 are to be arranged arranged in 6 ways to make numbers as 14532,15432,41532,45132,51432,54132

Thus no. of ways in which 5 digits no. divisible by 4 can be formed is 6+6=12

Is this solution Helpfull? Yes (3) | No (12)  

kushal jain  (12 years ago)

24.... because a no is divisible by 4 if its last two digits are divisible by 4 n the no of combination will be 24,12,32,52 and in them first 3 digit can be arranged in 3 factorial ways..... so total 3!*4=24 ans

Is this solution Helpfull? Yes (2) | No (4)  

K BASHA  (12 years ago)

what about 12,52 also divisible by 4

then the ans not 12 check once Mr rakesh

Is this solution Helpfull? Yes (1) | No (4)  

Rakesh Kumar Kohli  (12 years ago)

Ans. 24
For divisible by 4 the last two digits of the number should be divisible by 4.
So the numbers must have last two digits as 24,32,12 or 52.
If last two digits are 24 then digits 1,3,5 are to be arranged arranged in 6 ways to make numbers as 13524,15324,31524,35124,51324,53124
If last two digits are 32 then digits 1,4,5 are to be arranged arranged in 6 ways to make numbers as 14532,15432,41532,45132,51432,54132.
In the same way we can make 6 numbers with last two digits as 12 and 52 respectively.
Thus no. of ways in which 5 digits no. divisible by 4 can be formed is 6*4 = 24

Is this solution Helpfull? Yes (1) | No (4)  

PRIYADHARSHINI P  (9 years ago)

Last two digits can be 12,24,32,44,52.... 5 combinations

1 st digit can be from 1 to 5 so it can be filled in 5ways

similarly 2nd digit and 3rd digit also 5 ways
5*5*5*5=625

THEN 5 combinations so 625*5=3125

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

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