Latest IBM Aptitude Question SOLUTION: How many 3 digit odd number can be formed which are divisible by 5 using the numbers from 0 to 9(without repetition).

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How many 3 digit odd number can be formed which are divisible by 5 using the numbers from 0 to 9(without repetition).

Asked In IBM (12 years ago)

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shivani singh  (7 years ago)

If the no is divisible by 5 then the last digit must be 0 or 5.
But since no has to be odd so last digit will be only 5(fixed).
First digit can be 1,2,3,4,6,7,8,9=8digits
Second digit can be 0 also so there can be 8digits.
ans=8*8*1=64

Is this solution Helpfull? Yes (16) | No (3)  

Rimu  (10 years ago)

ans 56
8n2=56

Is this solution Helpfull? Yes (6) | No   

SAIALEKHYA  (10 years ago)

_ _ _
(1)(2) (3)
(3)--> 2 ways i.e 0,52
(2)--> 9
(1)-->8
therefore 2*9*8 ways= 144

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

Shikha Devi  (11 years ago)

ans: 12
three digit odd no from 0-9 are( without repetition)
135,175,195
315,375,395
715,735,795
915,935,975 so total no are 12

Is this solution Helpfull? Yes | No (12)  

Rishav Dutta  (10 years ago)

last two digits must be 0, 5 so that it can be divisible by 5. So those two numbers can be arranged in 2p1 ways... i.e. 2
Rest two digits can be arranged in 10p2 ways... if rep is allowed.
Sp total combo: 10p2* 2p1=180 ans

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Subhankar Chowrasia  (7 years ago)

ans 100
(5,15,25,35,45,55,65,75,85,95)= 10*10=100 numbers

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md wasi ahmed ansari  (7 years ago)

0,1,2,3,4,5,6,7,8,9=total 10 digits

_,_,_
3rd place an be filled by=1/3/5/7/9 so total=5 ways
1st place 0 cant be there coz it will be 2 digit no.
so excluding last digit we have (10-1)=9 digit remains
so 0 exclude remaining=8 digits
so 1st place can takes 8 digit and 2nd place can take 7 digit(in this case it takes 8 coz 0 in the middle is allowed)
ans =8*8*5=320
Enjoy

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SraVani  (5 years ago)

There are three places _ _ _
Now the first two places can be anything so 9p2
Now the last one can be only 5,0 because divisble by 5 so 2c1
So ans is 5c2*9p2

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