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

• 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
• 10 years agoHelpfull: Yes(43) No(3)
• 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
• 10 years agoHelpfull: Yes(12) No(0)
• 11111*(1+3+5+7+9)*4!
  11111*(25)*24
  6666600
  
• 10 years agoHelpfull: Yes(9) No(0)
• 6666600 is the ans.....
• 10 years agoHelpfull: Yes(3) No(1)
• Logic is (n-1)P(r-1)*(111111.......r times)*(sum of digits)............n=number of digits given,r=r digit number
• 10 years agoHelpfull: Yes(2) No(1)
• its so simple when ever u c dis type
  go for it
  11111(1+3+5+7+9)*(5-1)!=6666600
• 10 years agoHelpfull: Yes(2) No(0)
• PALLAVI PLZZ GIVE UR NO........:)
  
• 9 years agoHelpfull: Yes(2) No(0)
• @ SAUMYADIP MOITRA plz plz explain how????????
• 10 years agoHelpfull: Yes(1) No(2)
• 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
• 10 years agoHelpfull: Yes(1) No(0)
• 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?
• 10 years agoHelpfull: Yes(0) No(1)
• 5!(11111)(25)
• 10 years agoHelpfull: Yes(0) No(3)
• 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
• 10 years agoHelpfull: Yes(0) No(2)
• 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
• 10 years agoHelpfull: Yes(0) No(0)
• 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
  
• 9 years agoHelpfull: Yes(0) No(0)
• sum=sum of digit*(d-1)!*11111
  sum=25*(5-1)!*11111
  sum=25*24*11111
  sum=6666600
• 9 years agoHelpfull: Yes(0) No(0)
• ans: sum=666660
• 9 years agoHelpfull: Yes(0) No(0)
• sum=sum of digit*(d-1)!*11111
  sum=25*(5-1)!*11111
  sum=25*24*11111
  sum=6666600..is correct.lock kr do
• 9 years agoHelpfull: Yes(0) No(0)