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

Asked In TCS Nakshatra Mukhopadhyay (8 years ago)

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Souradeep Guhathakurta  (8 years ago)

last 2 digits shud be divisible by 4..possible pairs are 12,24,32,44,52...the other 7 places have 5 possibilities. therefore.. 5^7*5==390625

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jithin prakash  (8 years ago)

5^8*2=781250
here the rule for division by 4 is ,last two digits divisible by 4,so we have to take 12,24,32,44,52 in the last two digits.hence we can see in 10 th position also contain 1,2,3,4,5 number ,only the 1st number should be 2 or 4.so in 9 digit number first digit should be 2 or 4 ,all other position have 1,2,3,4,5,
so total number is=5*5*5*5*5*5*5*5*2=781250

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Shashank Sourabh Singh  (8 years ago)

last two digit may be 12,24,32,44,52
so 5^7*5=390625

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Golluru Lalitha  (8 years ago)

_ _ _ _ _ _ _ _ _
5 5 5 5 5 5 5
12
24
32
44
52
For divisible by 4 last 2 digits should be divisible 4. their are 5 possibilities, and remaining each digit having 5 possibilities. for 7 digits it becomes 5^7.
.". ans= 5^7*5 = 390625

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waivaw upadhyay  (8 years ago)

(5^7*3*1 + 5^7*2*1)
this is because last two digits should be among 12,24,32,44,52...and we see carefully,it is quite evident that the last digit will be 2 if the second last digit is an odd number and 4 if second last digit is an even number.

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