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 seperating 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.
  
  ANS. B) 1
• 9 years agoHelpfull: Yes(36) No(1)
• b,11(y+11x+121w+1331v)+z=11*13727+4,z=4
  11(x+11w+121v)+y=11*1247+10,y=10
  11(w+11v)+x=11*113+4,x=4
  11v+w=11*10+3,v=10,w=3
  so it gives the unique sol
  
• 9 years agoHelpfull: Yes(3) No(0)