the sum of all possible two digit numbers formed from threee different one digit natural numbers when divided by the sum of original three no.s is equal to?

• Let the three nos. be x,y,z
  So,the sum of all two digit nos formed by them is 10x+y+10x+z+10y+x+10y+z+10z+x+10z+y=22(x+y+z)
  So22(x+y+z)/(x+y+z)=22
• 11 years agoHelpfull: Yes(25) No(3)
• sorry th correct answer is =33
  i misunderstood the question
  take p,q,r now 3*3=9 no will be formed
  sum of it (10*p+p)+(10*q+q)+(10*r+r)+(10*p+q)+(10*p+r)+(10*q+r)+(10*q+p)+(10*r+p)+(10*r+q)=33(p+q+r) now divide it by p+q+r u get 33
• 11 years agoHelpfull: Yes(6) No(3)
• The maximum sum of three different single digit numbers can not be greater than 24(9+8+7).
  Hence 36 is rejected.
  22 has factors, 1, 2 and 11only, whose sum is only 14; hence this also is not possible.
  The single digit factors of 18 are, 1, 2, 3, 6, 9. Of these sum of 3 + 6 + 9 = 18; Hence '18' is the required number for the given question.
  Source(s):
  My knowledge
• 11 years agoHelpfull: Yes(1) No(0)
• ANS: 22
  (10X+Y)+(10X+Z)+(10Y+X)+(10Y+Z)+(10Z+X)+(10Z+Y)=[20(x+y+z)+2(X+Y+Z)]/(X+Y+Z)
  =>22
  
• 11 years agoHelpfull: Yes(1) No(2)
• ans=6
  take any three digit let p,q,r
  total numbers formed of 2 digits =3*3=9
  total digits used in 9 numbers=9*2=18
  out of which 6 of p 6 of q and 6 of r now adding
  we get 6*(p+q+r).divide it by p+q+r u get 6
• 11 years agoHelpfull: Yes(0) No(4)