Latest TCS Aptitude Question SOLUTION: Alok is attending a workshop how to do more with less and todays theme is working with fewer digits. The speakers discuss how a lot of miraculous mathematics can be achived if mank

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Alok is attending a workshop how to do more with less and todays theme is working with fewer digits. The speakers discuss how a lot of miraculous mathematics can be achived if mankind (as well as womankind) had only worked with fewer digits. The problem posed at lthe end of the workshop is how many 6 digit numbers can be formed using the digits 1, 2,3,4,5 (but with repetition) that are divible by 4? Can you help Alok find the answer?
3125
3906
1250
3907

Asked In TCS Aashu (12 years ago)

Unsolved Read Solution (5)

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Read Solution (5) :

HARINATHAREDDY  (12 years ago)

ans is 3125 because divisible rule for 4 is last two digits divisible then it divisible by 4. now last two digits are filled 5 ways only they are 12,24,44,32,52
so tot last two places are arrnage in 5 ways
5*5*5*5*5=3125

Is this solution Helpfull? Yes (6) | No (5)  

soumya goyal  (12 years ago)

- - - - - - here on last two places we can put 44, 12, 24, 32, 52 (all r divisible by 4). remaining first 4 places can be occupied by 1,2,3,4,5. since dere is repetition so for 1st place 5 ways are possible..same with 2nd place so 5*5. all four places can have any of the five nos..therefore 5*5*5*5. and lastly the selected combinations..hence 5^5=3125.

Is this solution Helpfull? Yes (5) | No (2)  

raj  (12 years ago)

According to me answer will be- 3125

Lets mark 6 places for 6 digits..
- - - - - -
Now at 1st place the no. of ways of selecting 1 digit from availalbe 5 digits is 5C1
As it is given Repeatation is allowed.

For 2nd place it is again 5C1 as repitaion is allowed.
similarly for 3rd and 4th places...
5c1 5c1 5c1 5c1 - -
Now for last 2 places in order to be divisible by 4....last 2 places must always end with 12, 24, 32, 44, 55 (you can see it by writing the counting table of 4= 4, 8, 12, 16...)
So No. of ways of choosing 1 combination from 5 combinations is 5C1

So finally, No of digits divisible by 4 are= 5C1*5C1*5C1*5C1*5C1
= 5*5*5*5*5
= 3125

Is this solution Helpfull? Yes (4) | No   

geetha  (12 years ago)

3125 because 5^5=3125

Is this solution Helpfull? Yes (3) | No (4)  

raj  (12 years ago)

According to me answer will be- 1250

Lets mark 5 places for 5 digits..
- - - - -
Now at 1st place the no. of ways of selecting 1 digit from availalbe 5 digits is 5C2
As it is given Repeatation is allowed.

For 2nd place it is again 5C2 as repitaion is allowed.
similarly for 3rd and 4th places...
5c1 5c1 5c1 5c1 -
Now for last place in order to be divisible by 4....last place must always end with 2 or 4 (you can see it by writing the counting table of 4= 4, 8, 12, 16...)
So No. of ways of choosing 1 digit from 2 digits (i.e 2 & 4) is 2C1

So finally, No of digits divisible by 4 are= 5C1*5C1*5C1*5C1*2C1
= 5*5*5*5*2
= 1250

Is this solution Helpfull? Yes | No (2)  

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