que:Find the sum of all the nos which can be formed by 2,4,6,8(repetitions not allowed)
ex:2+4+6+8+....24+.....68+.....468+

• for single digit no arranggement sum =20
  
  total 2 digit no possible =12 and sum is 12/2*(24+86)=66
  total for 3 digit no possible=4*3*2 =24 and sum is 24/2*(246+864)=13320
  total for 4 digit no possible=4*3*2*1=24 and sum is 24/2(2468+8642)=133320
  
  hence sum is 20+660+13320+133320= 147320 ans.
  
  
• 10 years agoHelpfull: Yes(44) No(5)
• EX=4*3*2*1
  24/4=6(4 is total number)
  .>
  6*(2+4+6+8)=120,
  120*1+120*10+120*100+120*1000,
  120*(1111)=133320
  
  
• 10 years agoHelpfull: Yes(12) No(9)
• 20+660+13320+133320=147320 ans
• 10 years agoHelpfull: Yes(12) No(2)
• formula-
  (n-1)!(sum of digit)*(1111)
  (3)!*20*1111
  =133320
• 10 years agoHelpfull: Yes(10) No(0)
• 2 + 8642 = 8644/2 = 4322
  
  4322 * 64 =276608 is ans.
• 10 years agoHelpfull: Yes(2) No(1)
• 64 no. can be formed using 2,4,6,8
  
  take min no. that is 2.. and largest no. that is 8642 and then make average of this that is 4322 and multiply it with 64.. u will find the sum of the numbers
• 10 years agoHelpfull: Yes(2) No(1)
• ans is 147320
• 10 years agoHelpfull: Yes(2) No(1)
• kumar vikash...your ans is correct....it was asked on 28 sept...but will u tell me plzz in which option was right....it is option (a).....actually i forgot on which option 147320 was written...is it option a or option b....my set was D ...plz reply...
• 10 years agoHelpfull: Yes(2) No(0)
• total number of single digit number = 4, their sum = (2 + 4 + 6 + 8) = 20
  total number of two digit number = 12 and sum of the digits at units place (2 * 3 + 4 * 3 + 6 * 3 + 8 * 3) = 60
  sum of two digits number = 60 + 60*10 = 660
  total number of three digits number = 4 * 3 * 2 = 24
  sum of digits at units place in all the three digits number = (2 + 4 + 6 + 8) * 6 = 120
  sum of the three digits numbers = 120 + 120 * 10 + 120 * 100 = 13320
  total 4 digit number = 24
  sum of all the 4 digits number = 13320 + 120 * 1000
  total sum = 147320
• 10 years agoHelpfull: Yes(2) No(0)
• can u please explain the exact logical reason???@harmanpreet singh
• 10 years agoHelpfull: Yes(0) No(0)
• sum=s1=(2+4+6+8+24+............+8462) total 64 terms
  s2=(8642+8462+..................+2) total 64 terms
  s1+s2=(8642+2)*64
  sum=s1=s2=(s1+s2)/2=276608
  
• 10 years agoHelpfull: Yes(0) No(2)
• For 1 digit max number= 8 and min 2 and avg= 5 * 4(4 is the total number of 1 digit )= 20.
  For 2 digit min 28 max 82 avg =50 total number of 2 digit= 12
  So 50*12=600
  For 3 max 862 min 246 avg 500 * 24 ( 24 is total number of 3 digit
  = 12000
  And for 4 digit it will be 120000 now add 120000+12000+600+20
• 10 years agoHelpfull: Yes(0) No(1)
• kumar vikash how did u solve the que???
  
• 10 years agoHelpfull: Yes(0) No(1)
• total number of 1 digit number can be=2,4,6,8 ie 4
  total number of 2 digit number can be(repetition not allowed)=4*3=12 ways
  total number of 3 digit number can be(repetition not allowed)=4*3*2=24 ways
  total number of 4 digit number can be(repetition not allowed)=4*3*2*1=24 ways
  now smallest number can be 2
  largest number can be 8642
  so average of smallest and largest will be=8642+2/2=4322
  total selections of 1,2,3,4 digit combinations=4+12+24+24=64
  so sum wil be=8644*64=276608
• 10 years agoHelpfull: Yes(0) No(6)