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(#M40012757) TCS question Keep an EYE

Alok is attending a workshop "How to do more with less" and today's theme is “Working with fewer digits”. The speakers discuss how a lot of miraculous mathematics can be achieved if mankind (as well as womankind) had only worked with fewer digits. The problem posed at the end of the workshop is “How many 7 digit numbers can be formed using the digits 1, 2, 3, 4, 5 (but with repetition) that are divisible by 4?” Can you help Alok find the answer?

Asked In TCS ankita khare (11 years ago)

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snehil  (11 years ago)

last 2 digits we can hv only 12, 24,32,44 or 52 (since divisible by 4),
5 place 5 digits, way of arrange 5^5
ans will be 5*5^5=5^6

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AS  (11 years ago)

ans is 5^6

Is this solution Helpfull? Yes (29) | No (4)  

preeti arya  (11 years ago)

4*5^5
5^5 for first 5 digits and at the last 2 digits we can hv only 12, 24,32 or 52 (since divisible by 4), der r 4 ways 2 arrange, so ans is 4*5^5.

Is this solution Helpfull? Yes (3) | No (7)  

taruna  (11 years ago)

ans : 5^6.

Is this solution Helpfull? Yes (2) | No (1)  

ravindra joshi  (11 years ago)

ans is 5^6

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preeti arya  (11 years ago)

sory..... it was 5^6 coz 44 should also be considered.

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shubham malhotra  (9 years ago)

here we have to develop those no. which are dividable by 4. For this last two digit no must be dividable by 4. That can only be in this case is (12,24,32,44,52)

now we have to design the starting five digit that can be taken as

5p5 that is 5*4*3*2*1 = 120

now total no. that can be formed is 120*5 = 600.

so Ans is 600.

Thanks

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bhawesh kumar  (11 years ago)

please read the question correctly
wrong question!
7 digit number can not be formed using 5 digits only.

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