Latest Elitmus Aptitude Question SOLUTION: W*3^3 + X*3^2 + Y.3 + z = 117 What are the possible solution are there??

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(#M40034172) ELITMUS Verbal Ability question Number System Keep an EYE

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

Asked In Elitmus tausif (10 years ago)

Unsolved Read Solution (11)

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SHIVANI  (10 years ago)

(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)

Is this solution Helpfull? Yes (48) | No (3)  

Akanchha Raj  (10 years ago)

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

Is this solution Helpfull? Yes (16) | No (1)  

arpit jain  (10 years ago)

since it is not given that only integral solutions are required......hence infinite is the answer

Is this solution Helpfull? Yes (11) | No (2)  

tausif  (10 years ago)

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

Is this solution Helpfull? Yes (3) | No (1)  

Roopa Gnanashekar  (10 years ago)

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)

Is this solution Helpfull? Yes (3) | No   

suryam surada  (10 years ago)

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

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swati  (10 years ago)

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

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komal pandhariwal  (10 years ago)

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

Is this solution Helpfull? Yes (1) | No (1)  

Arpit Agrawal  (10 years ago)

@ 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.

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priyanka   (10 years ago)

can u tell me from where i should prepare for the amcat & elitmus..tausif

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Pooja baghel  (9 years ago)

ans a is correct

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