If v, w,x,y,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

• 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
• 7 years agoHelpfull: Yes(16) No(1)
• v(11^4)+w(11^3)+x(11^2)+y(11)+z= 151001
  
  => 11*( v*11^3 + w*11^2+ x*11 + y ) + z = 11*13727 + 4
  
  => ( v*11^3 + w*11^2+ x*11 + y )= 13727 & z = 4
  
  => 11*( v*11^2 + w*11 + x ) + y = 11*1247 + 10
  
  => ( v*11^2 + w*11 + x )= 1247 & y = 10
  
  => 11*( v*11 + w ) + x = 11*113 + 4
  
  => ( v*11 + w )= 113 & x = 4
  
  => ( v*11 + w ) = 11*10 + 3
  
  => v = 10 & w = 3
  
  so, v=10, w=3, x=4, y=10 ,z=4
  
  (v,w,x,y,z)=(10,3,4,10,4)
  
  so, only one combination
• 7 years agoHelpfull: Yes(10) No(1)
• refer this link for ans:
  https://www.algebra.com/algebra/homework/Permutations/Permutations.faq.question.857977.html
• 7 years agoHelpfull: Yes(5) No(0)
• how we are getting quotient 10 and remainder 4591 ?
• 7 years agoHelpfull: Yes(0) No(3)
• Answer is 1
• 1 year agoHelpfull: Yes(0) No(1)