find the sum of the all the number formed by 2,4,6, and 8 without reputation. number may be of any digit like 2, 24, 684, 4862.

• single digit no formed = 2,4,6,8
  sum= 2+4+6+8= 20
  two digit= 24,26,28,42,46,48,62,64,68,82,84,86
  sum= 660
  three digit no= there are possiblity of three like all possiblity from 246,268,468,248
  sum of all possiblity=(n-1)!*(sum of the n digits)*111)
  using the above formula--
  sum of 246 possiblity = 2*12*111=2664
  sum of 268 possiblity = 2*16*111=3552
  sum of 468 possiblity = 2*18*111= 3996
  sum of 248 possibility = 2*14*111= 3108
  
  sum of all 4 digit nos.= 6*20*1111=(atleast use that much of brain of urself..:)
  
  total sum= 147320
  sum=
• 9 years agoHelpfull: Yes(29) No(8)
• single digit(one) sum of digit is=(2+4+6+8)=20-------------(1)
  sum of two digit no is = 11(two digit) *( no of times digit repeat )* (sum of all digit)
  =11*3*(2+4+6+8)
  =11*3*20 =660 --------------(2)
  sum of three digit no is =111*6*20
  =13320 ----------------------(3)
  sum of four digit no is = 1111*6*20
  = 133320 --------------------(4)
  
  adding eqn 1,2,3 and 4
  we get sum of all no by forming digit (total sum) =147320
• 9 years agoHelpfull: Yes(18) No(3)
• 2+4+6+8=20
  24+26+28+46+48+68=240
  42+62+82+64+84+86=420
  246+264+426+462+624+642=2664
  268+286+628+682+826+862=3552
  248+284+428+482+824+842=3108
  468+486+648+684+846+864=3996
  2468+2486+2648+2684+2846+2864=15996
  4268+4286+4628+4682+4826+4862=27552
  6248+6284+6428+6482+6824+6842=39108
  8246+8264+8426+8462+8624+8642=50664
  so,20+240+420+2664+3552+3108+3996+15996+27552+39108+50664=147320
• 10 years agoHelpfull: Yes(12) No(19)
• 4+12+24+24=64
  
• 10 years agoHelpfull: Yes(3) No(20)
• single digit no = 4
  2 digit no= 12
  3 digit no= 24
  4 digit no =24
  
  total for the 1 digit no = (2+4+6+8 =20)
  total for the 2 digit no => sum of digits the unit's place + sum of digits at ten's place
  sum of the digits in the unit's place =>
  as each of the 2,4,6,8 will repeat 12/4 =3 times in at units place then sum = (3*20 =60)
  similarly sum of the digits at ten's place =60
  so total of 2 digit no =60(10^0 +10^1)=660
  total of 3 digit no = sum of digits at unit'place + sum of digits at tens place +sum of digits at hndrd's place
  sum of digits at unit's place =>
  as each of 2,4,6,8 will repeat 24/4=6 times at units place then sum =(6*20=120)
  similarly at the tens and hundred place it will repeat 6 times so sum =120
  total of 3 digit no =>120*(10^0 +10^1+10^2)=13320
  similarly total of 4 digit no =>120*(10^0 +10^1+10^2+10^3) =133300
  sum of all no's => 20+660+13320+133320 = 147320
  
• 9 years agoHelpfull: Yes(3) No(0)
• This question can be solved by using a formula.
  1. add the given numbers
  here, 2+4+6+8= 20
  2. Multiply the resultant by no. of 1s each for a digit
  for example for 2( one digit)- it will be multiplied by 1
  2,4( two digits)- it will be multiplied by 11
  2,4,6( three digits) it will be multiplied by 111
  2,4,6,8( four digits) it will be multiplied by 1111
  3. now, multiply the resultant by no. of times any digit appears in any position.or u can say no. of times a digit will repeats itself at a fixed position
  1 digit can be placed in one position in one way only
  for 2 digits- there can be 2 positions_ _ if i fix any digit at any place then other positions can be filled in 3 ways.
  for 3 digits - there can be 3 positions _ _ _ if i fix any digit at any place then other positions can be filled in 3*2= 6 ways ( as there are 4 digits given )
  for four digits there are 4 position _ _ _ _ if i fix any one digit at any one place others can be filled in 3*2*1 =6
  
  now ,
  1. for 1 digit like 2= 20*1*1= 20
  2. for 2 digits like 24= 20*11*3 =660
  3. for 3 digits like 684 = 20*111*6=13320
  4. for 4 digits like 4862= 20*1111*6= 133320
  so the answer= 133320+13320+660+20= 147320
  
  
  
  
  
  
  
• 9 years agoHelpfull: Yes(2) No(0)
• wrong question
• 10 years agoHelpfull: Yes(1) No(16)
• single digit no. will be 4
  2 digit no. will be 12
  3 digit no. will be 24
  4 digit no. will be 24
  so total will be 64
• 9 years agoHelpfull: Yes(1) No(8)
• ans is 133320.there is a direct formula. (n-1)!*sum of the digits*1111......continue till n terms.it is vaid wen repeatation is not allowed.
• 9 years agoHelpfull: Yes(1) No(6)
• its formula is n-1!(sum of digits)*(1111....)ntimes
  in this qstn n=4 so n-1 =3
  sum of digits = 20
  
  
  
  actually the digits are 2,4,6,8 ,, sum 20
  lets cum frm thousands place so _ _ _ _ in this let 2 be in thousands place then rest can be arranged in
  3! ways.. therefore it is 2000*3!
  like wise formula can be 3!(20)(1111)= 133320
• 9 years agoHelpfull: Yes(1) No(1)
• 4!=24
  so
  sum of numbers=n*(n+1)/2=24*23/2=276
• 10 years agoHelpfull: Yes(0) No(8)
• 2+4+6+8=20for single digit numbers
  (2+4+6+8)(3p1*11)=660 sum for double digit number
  20*3p2*111=13320 sum for triple digit numbers
  20*3p3*1111=22220 sum for four digit numbers
  so and is 20+660+13320+22220=36220
• 10 years agoHelpfull: Yes(0) No(5)
• this is for 4 digit number
  6*(2+4+6+8)*1000
  6*20*100
  6*20*10
  6*10*1
  total sum =133320
  for 3 digit number
  6*20*100
  6*20*10
  6*20*1
  total sum=13320
  for 2 digit number
  6*20*10
  6*20*1
  total sum=1320
  for 1 digit
  6*20*1=120
  total sum=133320+13320+1320+120=148080(**might have calculation error but method is correct**)
  
• 8 years agoHelpfull: Yes(0) No(0)
• If 4 digit number is given then i have simple trick...
  1111 * 6*(2 + 4 + 6 + 8)
  
  
  suppose problem is given 3 numbers like 2,4 and 8...then formula is :
  111 * (the number of times any digit appears in any position) * (2 + 4 + 8)
• 7 years agoHelpfull: Yes(0) No(0)