if v,w,x,y,z are non negative intergers each leass than 11,then how many distinct combinations are possible of(v,w,x,y,z) satisfy v(11^4) +w(11^3)+ x(11^2)+ y(11) +z=151001

• change151001 to base 11 number .
  it will be
  10 3 4 10 4
  
  v(11^4) +w(11^3)+ x(11^2)+ y(11) +z=151001
  where v=10, w=3,x=4, y=10, z=4
  
• 12 years agoHelpfull: Yes(94) No(6)
• 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 only ONE solution.
• 10 years agoHelpfull: Yes(31) No(1)
• Only 1 possible combination.
  
  =10, w=3,x=4, y=10, z=4
• 12 years agoHelpfull: Yes(20) No(2)
• 11(11^3*v + 11^2*w + 11*x + y) + z = 11*13727 +4
  so z=4
  11(11^2*v + 11*w + x ) + y = 11*1247 +10
  so y=10
  11(11*v + w ) + x = 11*113 +4
  so x=4
  11*v + w = 11*10 + 3
  so w=3
  v=10
• 10 years agoHelpfull: Yes(14) No(1)
• if v,w,x,y,z are non negative intergers each leass than 11,then how many distinct combinations are possible 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
• 12 years agoHelpfull: Yes(11) No(11)
• i think its asking 4 distinct combination of v,w,x,y,z so the answer will be
  0
  
• 12 years agoHelpfull: Yes(5) No(11)
• 11^4v+ 11^3w + 11^2x +11^y +z = 151001
  11(11^3v+11^2w+11^x+y) + z = 11*13727 + 4 so, z = 4
  now,
  11(11^3v+11^2w+11^x+y)=11*13727
  11(11^2v + 11^w +x) + y =11*1247 + 10 y = 10
  
  11(11^v + w ) + x = 11*113 + 4 x = 4
  
  11^v + w = 11*10 + 3 v=10 , w=3
• 9 years agoHelpfull: Yes(2) No(0)
• how u can said only one combinatio is possible
• 11 years agoHelpfull: Yes(1) No(0)
• 0 Will be the answer as it is asking for distinct combinations.There are 2 10s in the solution provided.
• 11 years agoHelpfull: Yes(1) No(1)
• First of all, I think you should type the given equation as
  33 w + 32 x + 3y + z= 34,
  Since w & x are non-negative integers,
  both w and x cannot be greater than 0.
  In other words, one of w and x must be 0.
  When w =0, we have 32 x + 3 y + z = 34.
  If x = 0,then 3y+z = 34.
  But both y & z are less than 3, so 3y + z
• 10 years agoHelpfull: Yes(0) No(2)
• 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:
  
  ---------------------------------------------------
  
  So we show that that is the ONLY solution.
  
  v cannot be larger than 10.
  Suppose that v could be less than 10, then the largest
  
  v(11^4)+w(11^3)+x(11^2)+y(11)+z
  
  could possibly be would be when v=9, w=10, x=10, y=10, z=10
  
  9(11^4)+10(11^3)+10(11^2)+10(11)+10 = 146409
  
  But that is less than 151001. So v can only be 10.
  
  So we know that:
  
  10(11^4)+w(11^3)+x(11^2)+y(11)+z=151001
  
  146410+w(11^3)+x(11^2)+y(11)+z=151001
  
  w(11^3)+x(11^2)+y(11)+z=4591
  
  If w were more than 3, the smallest
  
  w(11^3)+x(11^2)+y(11)+z
  
  could be is when w=4,x=0,y=0,z=0
  
  4(11^3)+0(11^2)+0(11)+0 = 5324
  
  But that is larger than 4591, so w is not greater than 4.
  
  On the other hand,
  
  If w were less than 3, the largest
  
  w(11^3)+x(11^2)+y(11)+z
  
  could be is when w=2,x=10,y=10,z=10
  
  2(11^3)+10(11^2)+10(11)+10 = 3992
  
  But that is smaller than 4591, so w is not less than 3 either.
  
  Therefore w = 3 is the only possibility, and
  
  3(11^3)+x(11^2)+y(11)+z = 4591
  x(11^2)+y(11)+z = 598
  
  
  If x were more than 4, the smallest
  
  x(11^2)+y(11)+z
  
  could be is when x=5,y=0,z=0
  
  5(11^2)+0(11)+0 = 605
  
  But that is larger than 598, so w is not greater than 4.
  
  On the other hand,
  
  If x were less than 4, the largest
  
  x(11^2)+y(11)+z
  
  could be is when x=3,y=10,z=10
  
  3(11^2)+10(11)+10 = 483
  
  But that is smaller than 598, so x is not less than 4 either.
  
  Therefore x = 4 is the only possibility, and
  
  4(11^2)+y(11)+z = 598
  y(11)+z = 114
  
  Y cannot be more than 10.
  If y were less than 10, the largest
  
  y(11)+z
  
  could be is when y=9,z=10
  
  9(11)+10 = 109
  
  But that is smaller than 114, so x = 10.
  
  
  So we have proved that v=10, w=3, x=4, y=10, z=4
  
  is the only solution.
  
  
  Source : https://www.algebra.com/algebra/homework/Permutations/Permutations.faq.question.857977.html
• 8 years agoHelpfull: Yes(0) No(0)
• 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.
  1 Answer
• 8 years agoHelpfull: Yes(0) No(0)
• 151001/11^4 = 10 with remainder = 4591
  
  4591/11^3 = 3 with remainder = 598
  
  598/11^2 = 4 with remainder = 114
  
  114/11 = 10 with remainder = 4
  
  Therefore 151001=10(114)+3(113)+4(112)+10(11)+4
  
  i.e., (v,w,x,y,z) is (10,3,4,10,4). These values satisfies given conditions and is one valid option.
  
  Let's analyze if further option available satisfying the given condition. v,w,x,y,z should be less than 11 and greater than zero. The solution we got have v as 10. Suppose v=9. With this, the maximum value possible under the given constraints is
  9(114)+10(113)+10(112)+10(11)+10=146409
• 7 years agoHelpfull: Yes(0) No(0)
• answer is 1 because it is asking how many distinct combination. It is not asking for distinct values.
• 5 years agoHelpfull: Yes(0) No(0)