Latest TCS Aptitude Question SOLUTION: Alok is attending a workshop 'How to do more with less1 and today's theme is working with fewer digits. The speakers discuss how a lot of miraculous mathematics can be achieved if

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(#M40011008) TCS question Keep an EYE

Alok is attending a workshop 'How to do more with less1 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 8 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 (11 years ago)

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Nabeel Ansari  (11 years ago)

ans is 78125
the condition for divisibility by 4 is that last two digit of any number should be divisible by 4.
and we can easily see that from these 5 digits available last two digit of 8 digit number can be filled in 5 ways i.e 12,24,32,44,52.and as the repitition is allowed the remaining 6 digits can be filled in 5*5*5*5*5*5.
so total 8 digit number can be formed is 5^6(above)*5(12,24,32,44,52)
so ans is 78125

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Ashish Singh  (11 years ago)

for 8 digit no total total no form is 5^7(because last 2 place has only 5 conbination whic devide by 4)

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sushma  (11 years ago)

5^7
5^6 ways for the first 6 places and 5 ways for the last two places(12,24,32,44,52)
as for the number to be divided by 4 ,the last two digits must be divisible by 4

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Vimalesh  (11 years ago)

ans 5^7
no divisible by 4 .last 2 digit should be divided by 4..
possibility is 12,24,32,44,52..so one possibility.last two got placed...
remaining 6 digits.5^6 and 5(12,24,44,52,32)..
then 5^7

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palak shah  (11 years ago)

12
24
32
44
52
so for last two digits dere are total 5 possibilities
and for first six nos dere are 5 possibilities for each(coz repetition is allowed)
so total possibilities
5*5*5*5*5*5*5
=78125

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