If v,w,x,y and z are non negative integers, each less than 11, than how many distinct combination of (v,w,x,y,z) satisfy v(11^4) + w(11^3) + x(11^2) + y(11) + z=151001

• 11(11^3v+11^2w+11x+y) + z = 11x13727 + 4 [hence; z = 4]
  repeating the above procedure;
  11(11^2v + 11w +x) + y =11x1247 + 10 [y = 10]
  11(11v + w ) + x = 11x113 + 4 [x = 4]
  11v + w = 11x10 + 3 [v=10 , w=3]
  So one combination
• 9 years agoHelpfull: Yes(41) No(1)
• 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
• 9 years agoHelpfull: Yes(9) No(0)
• only one combinatin satisfy
  the value of v=10
  w=3
  x=4
  y=10
  z=4
  which is in base 11 form.
• 9 years agoHelpfull: Yes(2) No(0)
• bhai log maine elitmus ka exam 28 september ko diya tha kya hum phir se 5 oct.ko de sakte hai same id se ,or dono ka result declare hoga ya jo last me diye uska pls batao
  
• 9 years agoHelpfull: Yes(2) No(1)
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