If v,w,x,y, and z are non negative integers, each less than 11, then how many distinct combinations of (v,w,x,y,z) satisfy v(11^4) + w(11^3) +x(11^2) + y(11) + z =151001 ?
a)0
b)1
c)2
d)3

• Taking 11 common from LHS separating z and breaking 151001 as a factor of 11 with some remainder in RHS
  11(113v+112w+11x+y) + z = 11x13727 + 4 [hence; z = 4]
  Equating z =4 and cancelling 11 from each side we get the following simplified equation and again repeating the above procedure;
  11(112v + 11w +x) + y =11x1247 + 10 [y = 10]
  11(11v + w ) + x = 11x113 + 4 [x = 4]
  11v + w = 11x10 + 3 [v=10 , w=3]
  So we get the unique solution for the above equaton.
• 9 years agoHelpfull: Yes(27) No(3)
• b) 1
  v=10, w=3, x=4, y=10,z=4
  is the only solution
• 9 years agoHelpfull: Yes(3) No(2)
• 11(113v+112w+11x+y) + z = 11x13727 + 4 [hence; z = 4]
  Equating z =4 and cancelling 11 from each side we get the following simplified equation and again repeating the above procedure;
  11(112v + 11w +x) + y =11x1247 + 10 [y = 10]
  11(11v + w ) + x = 11x113 + 4 [x = 4]
  11v + w = 11x10 + 3 [v=10 , w=3]
  So we get the unique solution for the above equaton.
• 8 years agoHelpfull: Yes(0) No(0)