Latest TCS Aptitude Question SOLUTION: How many 9 digit numbers are possible by using 1,2,3,4,5 which are divisible by 4, if the repetition is allowed

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How many 9 digit numbers are possible by using 1,2,3,4,5 which are divisible by 4, if the repetition is allowed

Asked In TCS (12 years ago)

Solved Atul Read Solution (6)

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Read Solution (6) :

Atul  (12 years ago)

last 2 digits can be 12,24,32,44,52(5 ways)
first 7 places can be filled in 5^7 ways
Last 2 places can be filled in 5 ways
So total ways= 5^7 * 5
=5^8

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Mayur Jain  (12 years ago)

Itz 5^8=390625(short-cut method) itz jus simple yaar

First Place Probablity Having A Digit Is 5(like 1,2,3,4,5)
Since Repetation Allowed Second Place also implies The same so on till 7 location's but the last two digits shud be divisible by 4 and from those 2 locations we can 5 different combination's so the answer for this question is 390625

Actual Formulae :
5^8=5^7*5

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Saurabh Kumar  (12 years ago)

Division of 4 is sum of last two digit is divisible by 4 so last two digits is 53 35 31 13 22 44 so 6 and remainig 7 digit is 5^7 so totel no is 5^7 * 6=468750

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kanha shrivastava  (12 years ago)

Itz 5^8=390625
For the divisibility of 4 we consider last two location as one location.

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Pankaj  (12 years ago)

Division of 4 is sum of last two digit is divisible by 4 so last two digits is 53 35 31 13 22 44 so 6 and remainig 7 digit is 5^7 so totel no is 5^7 * 6=468750

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saalim jabir  (12 years ago)

Ans: 312500

last two digits can be : 12,24,44,52
so, 5*5*5*5*5*5*5*4=312500

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