1, 2, 3 and 4 can form 256 different four digit numbers. If digits repeated, two of them are 1111 and 1113. Then find the sum of 256 numbers.

a)711040 b)711000 c)711038 d)711042

• s=n/2(a+L)
  s= 256/2(1111+4444)
  s=711040
• 7 years agoHelpfull: Yes(49) No(7)
• 4^4=256 ways
  (1+2+3+4)/4=5/2
  256 *5/2*(1000+100+10+1)=711040
• 7 years agoHelpfull: Yes(14) No(1)
• As each digit of four digit number (abcd) can take 4 options (1, 2, 3, 4) total there can be 4^4 numbers.
  
  Obviously each digit (a, b, c, and d) will take the value of 1, 2, 3, 4 equal number of times, so each digit will take the value of 1, 2, 3, 4 - 444=43=64444=43=64 times: units digit will take the values of 1, 2, 3, 4 - 64 times and the same with tens, hundreds, thousands digits.
  
  So the sum would be 64*(1+2+3+4)+64*10*(1+2+3+4)+64*100*(1+2+3+4)+64*1000*(1+2+3+4)=64*10*(1+10+100+1000)=711040.
  
  
  Hope it helps.
• 7 years agoHelpfull: Yes(12) No(11)
• Sn = n/2 (a+L)
  =256/2 (1111+4444)
  =711042
• 7 years agoHelpfull: Yes(10) No(1)
• (N)^(n-1)*(sumofdigits)*(111....ntimes)
• 6 years agoHelpfull: Yes(3) No(0)
• (N)^(n-1)*(sumofdigits)*(111....ntimes)
  711040
• 5 years agoHelpfull: Yes(0) No(0)