Latest TCS Aptitude Question SOLUTION: All possible Five digit number can be formed of 1,3,5,7,9 without any repetition. What will be the sum of the all numbers

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All possible Five digit number can be formed of 1,3,5,7,9 without any repetition. What will be the sum of the all numbers

Asked In TCS Arpit Jain (9 years ago)

Unsolved Read Solution (17)

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Read Solution (17) :

RAKESH  (9 years ago)

total no. formed = 5! =120
24 no.s starting with each digit 1,3,5,7,9
sum of digits starting with 1,3,5,7,9 = 24*(1+3+5+7+9) = 24*25 = 600
this sum will be same for thousand, hundred, tens & unit places

sum of all 5 digit numbers formed using digits 1,3,5,7,9
= 10000*600 + 1000*600 +100*600 + 10*600 + 600
= 600*(10000+1000+100+10+1)
= 600 * 11111
= 6666600

use formula
(1111......n) *( sum of digits ) *(n-1)!
here n = 5

Is this solution Helpfull? Yes (43) | No (3)  

Mohd Salim  (9 years ago)

The total 5 digit numbers that can be formed with the given digits are 5*4*3*2*1=120 numbers.

Now, the numbers that will be formed with the given digits in ascending order would be 13579, 13597, 13759 ........and so on........till .... 97351, 97513, 97531.

Now if u add the smallest and the largest of the numbers formed with the given digits (i.e 13579+97531) their sum would be 111110.

Now is u add the 2nd number and the 2nd last number (i.e 13597+97513) their sum will also be ..... 111110.

similarly, if u add the 3rd number and the 3rd last number ( i.e 13759 + 97351) their sum would also be 111110.

so the answer would be 111110*60=6666600

Is this solution Helpfull? Yes (12) | No   

Saumyadip Moitra  (9 years ago)

11111*(1+3+5+7+9)*4!
11111*(25)*24
6666600

Is this solution Helpfull? Yes (9) | No   

pritika singh  (9 years ago)

6666600 is the ans.....

Is this solution Helpfull? Yes (3) | No (1)  

kiran  (9 years ago)

Logic is (n-1)P(r-1)*(111111.......r times)*(sum of digits)............n=number of digits given,r=r digit number

Is this solution Helpfull? Yes (2) | No (1)  

aditya jain  (9 years ago)

its so simple when ever u c dis type
go for it
11111(1+3+5+7+9)*(5-1)!=6666600

Is this solution Helpfull? Yes (2) | No   

SUVANKAR GHOSH  (9 years ago)

PALLAVI PLZZ GIVE UR NO........:)

Is this solution Helpfull? Yes (2) | No   

Pallavi Verma  (9 years ago)

@ SAUMYADIP MOITRA plz plz explain how????????

Is this solution Helpfull? Yes (1) | No (2)  

suraj patil  (9 years ago)

total no. formed = 5! =120
24 no.s starting with each digit 1,3,5,7,9
sum of digits starting with 1,3,5,7,9 = 24*(1+3+5+7+9) = 24*25 = 600
this sum will be same for thousand, hundred, tens & unit places

sum of all 5 digit numbers formed using digits 1,3,5,7,9
= 10000*600 + 1000*600 +100*600 + 10*600 + 600
= 600*(10000+1000+100+10+1)
= 600 * 11111
= 6666600

Is this solution Helpfull? Yes (1) | No   

vikash  (9 years ago)

A three digit number was divided successively in order by 4, 5 and 6 leaving out the remainders. The
remainders were respectively 2, 3 and 4. How many such three digit numbers are possible?

Is this solution Helpfull? Yes | No (1)  

neha kumari  (9 years ago)

5!(11111)(25)

Is this solution Helpfull? Yes | No (3)  

Vijay Menaria  (9 years ago)

If repition is not allowed then sum of all will be
[(n−1)!*(111...ntimes)*(Sum of the digits)] − [(n−2)!*(111...(n−1)times)*(Sum of the digits)]
[(5-1)!]*(11111)*(1+3+5+7+9) - [(5-2)!]*(1111)*(1+3+5+7+9)
6666600 - 166650
6499950

Is this solution Helpfull? Yes | No (2)  

ANUBHA VERMA  (9 years ago)

logic is: the give nos(i.e 5..1,3,5,7,9) * (nos-1 ,i.e ..5-1=4)! * (sum of all the digits i.e., 1+3+5+7+9)
=6666600

Is this solution Helpfull? Yes | No   

chandni kumari  (9 years ago)

total no. formed = 5! =120
24 no.s starting with each digit 1,3,5,7,9
sum of digits starting with 1,3,5,7,9 = 24*(1+3+5+7+9) = 24*25 = 600
this sum will be same for thousand, hundred, tens & unit places

sum of all 5 digit numbers formed using digits 1,3,5,7,9
= 10000*600 + 1000*600 +100*600 + 10*600 + 600
= 600*(10000+1000+100+10+1)
= 600 * 11111
= 6666600

Is this solution Helpfull? Yes | No   

Mrittika Bhaumik  (9 years ago)

sum=sum of digit*(d-1)!*11111
sum=25*(5-1)!*11111
sum=25*24*11111
sum=6666600

Is this solution Helpfull? Yes | No   

Deepanshu Verma  (8 years ago)

ans: sum=666660

Is this solution Helpfull? Yes | No   

Abhishek nkumar  (8 years ago)

sum=sum of digit*(d-1)!*11111
sum=25*(5-1)!*11111
sum=25*24*11111
sum=6666600..is correct.lock kr do

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