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 repetition of digits is allowed?

• Placement Papers
• /
• TCS
• /
• Logical Reasoning
• /
• Number Series

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#M40002158) TCS Logical Reasoning question Keep an EYE

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

Asked In TCS partha (12 years ago)

Solved Kodi Read Solution (11)

Is this Puzzle helpful?   (68)   (16) Submit Your Solution Number Series

Read Solution (11) :

Kodi  (12 years ago)

5^8
last two is divisable by 4 is condition.
the possibility is 5(12,24,44,32,52)
5*5*5*5*5*5*5*5

Is this solution Helpfull? Yes (62) | No (26)  

soujanya  (12 years ago)

5^N-1 is da formula

Is this solution Helpfull? Yes (48) | No (10)  

Anshuman  (11 years ago)

In order to be divisible by 4, the two least significant digits must be
12, 24, 32, 44, 52 (a total of 5 choices)
The other 7 digits can be any of the five allowed numbers..
The total number of possible number is
(5^7)*(5) = 5^8 = 390625..

Is this solution Helpfull? Yes (32) | No (6)  

Rimu Mukherjee  (12 years ago)

5^8
will be the ans

Is this solution Helpfull? Yes (14) | No (10)  

baidyanath  (12 years ago)

last 2 digit should be divisible by 4, for divisibility by 4
so last two can be 12 24 32 44 52
now out of left seven places each can be filled in 5 way

therefore ans=5*5*5*5*5*5*5*4

Is this solution Helpfull? Yes (9) | No (37)  

Ravikant  (9 years ago)

If repetition is allowed, first 7 places can be filled in 5^7 ways and last 2 places can be 12, 24, 32, 52, 44 i. e. 5 ways therefore answer is 5^8

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

harshit varshney  (11 years ago)

please explain it how the ans of it can be find out.

Is this solution Helpfull? Yes (1) | No (5)  

ragavendiran  (8 years ago)

If A number has to be divisible by 4, the last two digits must be divisible by 4. So possibilities are, 12, 24, 32, 44, 52. And the of the remaining 7 places, each place got filled by any of the five digits. So these 7 places got filled by 5 x 5 x .....(7 times) = 5^7 ways. So total ways are 5 x 5^7 = 5^8

Is this solution Helpfull? Yes (1) | No   

sujit sinha  (7 years ago)

The number which is divisible by 4 has 0,2,4,6,8 in its unit place.
so the unit place can be filled by 2 different ways , the first digit in 4 ways and rest 8 digits in 5 ways .
Total number of ways =4*5^8*2=3125000

Is this solution Helpfull? Yes | No   

Marshall Navis  (7 years ago)

5 ^ 8 * 2 =781250

Is this solution Helpfull? Yes | No   

sundeep kumar  (6 years ago)

last two digits must be divisible by 4. So possibilities are, 12, 24, 32, 44, 52.
7 places, are left to fill by any of the five digits in this way we get
5*5*5*5*5*5*5*5=5^8

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.