W*3^3 + X*3^2 + Y.3 + z = 117
What are the possible solution are there??

• (3^3)w+(3^2)x+(3^1)y+z=117= (3*39)+0.........(1)
  comparing both sides we get z=0
  then take out 3 from both sides we get eqn as :
  (3^2)w+3x+y=39=3*13...............................(2)
  again comparing both sides we get y=0
  now eqn becomes
  3w+x=13.................(3)
  from above eqn 13 can be written as : 13=(3*4)+1
  13=(3*3)+4
  13=(3*2)+7
  13=(3*1)+10
  all equal to (3*w)+x
  comparing all we get 4 solutions as (w,x,y,z)=(4,1,0,0),(3,4,0,0),(2,7,0,0),(1,10,0,0)
• 10 years agoHelpfull: Yes(48) No(3)
• 3*(w*3^2+x*3+y)+z=3*39+0
  z=0
  3*(w*3+x)+y=3*13+0
  y=0
  3*w+x=3*4+1
  x=1, w=4
  x=1, y=0, z=0, w=4
• 10 years agoHelpfull: Yes(16) No(1)
• since it is not given that only integral solutions are required......hence infinite is the answer
• 10 years agoHelpfull: Yes(11) No(2)
• Jus go through Arun Sharma for Quant
  and try to solve Cryptairthmatic Problem and Data sufficiency for Logic
  and One word Substitution and Parajumble and RC For verbal
  Best Of Luck
  @Priyanka
  
• 10 years agoHelpfull: Yes(3) No(1)
• W*3^3 + X*3^2 + Y.3 + z = 117
  3(w*3^2+x*3+y)+z=3(39)+0 so z=0
  3(w*3+x)+y=3*13+0 so y=0
  3w+x=4*3+1 so w=3 and x=1
  (w,x,y,z)=(4,1,0,0)
• 10 years agoHelpfull: Yes(3) No(0)
• here the possible values for W is 0,1,2,3,4...possible values for x is 0 to 13 : y is 0 39 and Z is 0 to 117
  1)w=0; (40,37,34,........7,4,1): ...total is 287
  2)w=1: (31, 28,..........7,4,1):....total is 176
  3)w=2: (22, 19, 16.........7,4,1).....total is 92
  4)w=3:( 13, 10, 7, 4, 1 )......total is 35
  5)w=4: (4,1).......total is 5
  
  grand total is 287+176+92+35+5=595
• 10 years agoHelpfull: Yes(2) No(3)
• Solution:
  
  114v+ 113w + 112x +11y +z = 151001
  
  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
• 10 years agoHelpfull: Yes(2) No(0)
• 117 when divided by 27 gives remainder 9
  We have 9 + 108 =117 so we are left with 108
  We have to divide it such that it is divisible by 3 9 nd 27 so if we write 3*3 + 2*9
  We are left with 81 which is 3*27 so w=3 x=2 y=3 and z=9
• 10 years agoHelpfull: Yes(1) No(1)
• @ shivani -your solution is very good,but it also involves more answers like (0,13,0,0).
  Also we can keep w as -5 and x=2.In this way we can find several solutions to this ques and in my opinion answer should be infinity.
• 10 years agoHelpfull: Yes(1) No(0)
• can u tell me from where i should prepare for the amcat & elitmus..tausif
• 10 years agoHelpfull: Yes(0) No(0)
• ans a is correct
• 9 years agoHelpfull: Yes(0) No(0)