Find the sum of all the four digit numbers that can be formed using the digits 2,3,5,9. No digit is to be repeated.

• formula:
  sum of digits*(n-1)!*(111....number composed of n digits)
  soln:
  19*3!*1111=126654
• 11 years agoHelpfull: Yes(65) No(2)
• Be careful that your number has four distinct digits, otherwise the counting gets more confusing.
  
  There are 4!=24 such numbers. Add them up, without carrying first.
  
  In each column, you will have each of the four digits appear six times. So each column, without carries, will add up to 6(a+b+c+d). So the sum is
  
  6(a+b+c+d)(1+10+100+1000)=
  6(a+b+c+d)*1111=
  6666(a+b+c+d).
  =6666(2+3+5+9)=6666(19)=126654
• 11 years agoHelpfull: Yes(33) No(4)
• answer:126654
  explanation:(n-1)!(sum of all the digits)(1111...n times)
  here n=4.so,(4-1)!(2+3+5+9)(1111)
• 11 years agoHelpfull: Yes(7) No(1)
• answer is 226654
• 11 years agoHelpfull: Yes(4) No(8)
• @PRIYA
  what is the solution if digit is to be repeated....?
  and what is formula for it...??
• 11 years agoHelpfull: Yes(4) No(0)
• There are 4! = 24 numbers that can be formed. Now in each column we have each digit 6 times. So keeping that in mind,
  For unit's place sum is: (6*2)+(6*3)+(6*5)+(6*9)=114 --> 4 at units place and 11 in carry
  For ten's place sum is: (6*2)+(6*3)+(6*5)+(6*9)+11=125 -->5 at ten's place and 12 in carry
  For hundredth place sum is: (6*2)+(6*3)+(6*5)+(6*9)+12=126 -->6 at hundredth place and 12 in carry
  For thousand place sum is: (6*2)+(6*3)+(6*5)+(6*9)+12=126 -->6 at thousandth place and 12 in carry
  So combining the result we get: 126654
• 11 years agoHelpfull: Yes(3) No(0)
• give the logic
  
• 11 years agoHelpfull: Yes(2) No(3)
• answer is [6*1000*(2+3+5+9)]+[6*100*(2+3+5+9)]+[6*10(2+3+5+9)]+(6*1*(2+3+5+9)]=126654
• 11 years agoHelpfull: Yes(2) No(2)
• sum of digits is asked so it sould be 19 * no. of ways ie 4! so ans is 19*24= 456
• 11 years agoHelpfull: Yes(2) No(6)
• incase they ask with repetition of numbers..wat is the solutio
• 10 years agoHelpfull: Yes(2) No(0)
• i think answer is 126654
• 11 years agoHelpfull: Yes(1) No(2)
• we can arrange 2,3,5,9 in 16 ways and sum of 2,3,5,9 is equal to 19
  therefore 16*19=304
• 11 years agoHelpfull: Yes(0) No(10)
• in each column digit will be repeated six times as per solution..how 6 times?
• 11 years agoHelpfull: Yes(0) No(0)
• it will be 456 only....bcz 1st the digit starts with 2 and arrangements in this is may be 2359 ,2395,2539,2593,2935,2953 here 6 possibilities and each sum is 19 and 19*4= 114 and the sum became 114*6 for the num starting with 2,3,5,9 and it will be 456
• 11 years agoHelpfull: Yes(0) No(2)
• 2+3+5+9=19
• 8 years agoHelpfull: Yes(0) No(1)
• given is 2,3,5,9
  then 4!=24 and which is divided by 4 will give 6
  6(1000+100+10+1)*(2+3+5+9)=6666(2+3+5+9)
  then the answer is 126654
• 8 years agoHelpfull: Yes(0) No(0)
• can any one tell me how to solve this if the digits can be repeated?
• 7 years agoHelpfull: Yes(0) No(0)