if v,w,x,y,z are non negative intergers each leass than 7,then how many distinct combinations are possible of(v,w,x,y,z) satisfy v(7^4) +w(7^3)+ x(7^2)+ y(7) +z=151001

• 151001=62*7^4+6*7^3+0*7^2+0*7^1+1
  v=62
  w=6
  x=0
  y=0
  z=1
  only these combinations can produce the result
  so unique solution
• 6 years agoHelpfull: Yes(1) No(3)
• Dividing 151001 by 11^4 give quotient 10 and remainder 4591
  
  So
  
  151001 = 10*11^4 + 4591
  
  Dividing 4591 by 11^3 gives quotient 3 and remainder 598
  
  So
  
  151001 = 10*11^4 + 3*11^3 + 598
  
  Dividing 598 by 11^2 gives quotient 4 and remainder 114
  
  So
  
  151001 = 10*11^4 + 3*11^3 + 4*11^2 + 114
  
  Dividing 114 by 11 gives quotient 10 and remainder 4
  
  So
  
  151001 = 10*11^4 + 3*11^3 + 4*11^2 + 10*11 + 4
  
  So v=10, w=3, x=4, y=10, z=4
  
  is ONE solution. But is that the only one? We will
  show that it indeed is the only solution:
• 6 years agoHelpfull: Yes(1) No(1)
• In quest 11 s der not 7 n for 11 der exists only 1 solution
• 6 years agoHelpfull: Yes(0) No(1)
• 151001=62*7^7+6*7^3+1*7^2+0*7+32
  v=62
  w=6
  x=1
  y=0
  z=32
• 6 years agoHelpfull: Yes(0) No(1)