Latest Elitmus Aptitude Question SOLUTION: A five digit number is formed using digits 1, 3, 5, 7 and 9 without repeating any one of them. What is the sum of all such possible numbers? 1.6666600 2.6666660 3.6666666 4.None

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(#M40038706) ELITMUS Numerical Ability question Keep an EYE

A five digit number is formed using digits 1, 3, 5, 7 and 9 without repeating any one of them. What is the sum of all such possible numbers?
1.6666600
2.6666660
3.6666666
4.None

Asked In Elitmus Pratik Keshri (9 years ago)

Unsolved Read Solution (7)

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Read Solution (7) :

sakshi singh  (9 years ago)

Total Combinations without repetation is 5!=120.
smallest no = 13579 largest no = 97531.
take avg. = ((13579+97531)/2)*120= 6666600
ie ans is 1.

Is this solution Helpfull? Yes (108) | No (1)  

Subhash Kumar  (9 years ago)

Ans: 6666600

Explanation:
Given digits are 1,3,5,7,9

I am writing the numbers are in ascending form.. from given digits, numbers should be:

97531 + 97513 + 97351 + ........+ 13759 + 13597 + 13579

Now we 'll add first and last term of above sequence and again we 'll add second and second last term and so on...
we 'll get
97531 + 13579 = 111110
97513 + 13597 = 111110
97351 + 13759 = 111110
..
..
Upto middle position of the above sequence.
Now we see every time we are getting 111110

now i have to find the how many possible number :
==> 5*4*3*2*1 = 120
there are 120 numbers
then the middle position is: 120/2 = 60

so, sum of all possible numbers are....
60*111110 = 6666600 Ans.

Is this solution Helpfull? Yes (27) | No (2)  

Aniruddha  (9 years ago)

4!(1+3+5+7+9+)*11111

Is this solution Helpfull? Yes (14) | No (4)  

Ritesh Siriya  (9 years ago)

Sum of arrangement of such numbers are given by:(n-1)!*n times 1*(sum of digits)
here n = 5 and (n-1)! = 4! = 24..
sum of (first 5 odd) digits = n^2 = 25
thus 24*11111*25 = 6666600..option 1)

Is this solution Helpfull? Yes (9) | No   

Vivek singh bhatnagar  (9 years ago)

ans is 1.6666600

total possible no without replacement 5!=120

how many time 1,3,5,7,9 occur at unit place
24 times
_ _ _ _ 1 (1*24 => 24)
_ _ _ _ 3 (3*24 => 72)
_ _ _ _ 5 (5*24 => 120)
_ _ _ _ 7 (7*24 => 168)
_ _ _ _ 9 (9*24 => 216)

24+72+120+168+216 = 600 {Extract '0' from 600 for sum of all possible no acting as a unit place..
so remain digit is 60.....Carry-1}

again how many time 1,3,5,7,9 occur at ten place
6 times
_ _ _ 1 (1*6 => 6)
_ _ _ 3 (3*6 => 18)
_ _ _ 5 (5*6 => 30)
_ _ _ 7 (7*6 => 42)
_ _ _ 9 (9*6 => 54)

6+18+30+42+54 => 150
now add 150 to carry 60
150+60 = 210...{Extract '0' from 210 for sum of all possible no acting as a Ten place..so remain digit is 21.....Carry-2}


doing in this way i got my ans 6666600.....

Is this solution Helpfull? Yes (4) | No (3)  

prasanthi  (9 years ago)

sum of all such possible numbers formula=n*(n-1)!*11111
here n=5=5*4!*11111=6666600

Is this solution Helpfull? Yes (2) | No (3)  

AFTAB ALAM  (6 years ago)

If you assume that any digit is in fixed position, then the remaining four digits can be arranged in 4! = 24 ways.

So, each of the 5 digit will appear in each of the five places 24 times. So, the sum of the digits in each position is 24(1+3+5+7+9) = 600

The sum of all such numbers will be 600(1 + 10 + 100 + 1000 + 10000) = 6666600

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