Latest Elitmus Aptitude Question SOLUTION: (x-y)^n-(y-z)^n+z^n always divisible by what? given x=yz. options: 1)yz nt rember other options.

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(#M40037874) ELITMUS Numerical Ability question Keep an EYE

(x-y)^n-(y-z)^n+z^n always divisible by what? given x=yz.
options:
1)yz
nt rember other options.

Asked In Elitmus Sonali Ghadai (9 years ago)

Unsolved Read Solution (10)

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Read Solution (10) :

Niranjan C  (9 years ago)

here its given: x=yz, so the equation simplifies to :
(yz-y)^n-(y-z)^n+z^n

now substitute n=1:
then we get (yz-y)-(y-z)+z == yz

now substitute n=2:
then we get (yz-y)^2-(y-z)^2+z^2 == (yz)^2+zy^2+2yz
so this is also divisible by yz

try 3,4 and so on....
the above equation would be divisible from yz.

hence ans:1)yz

Is this solution Helpfull? Yes (9) | No (16)  

deepak  (9 years ago)

option
a)yz
b)z but not y

answer is b) z but not y

Is this solution Helpfull? Yes (5) | No (6)  

deepak  (9 years ago)

@ rakesh can u solve this???

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

Pralay Saha  (9 years ago)

x=yz so(yz-y)^n-(y-z)^n+z^n={(yz)^n-......+(y)^n}-{(y)^n-.......+(z)^n}+(z)^n
=(yz)^n-.......+(y)^n-(y)^n+.........-(z)^n+(z)^n
If we solve this equation we see the result is always divisible by yz...
so the answer is option 1)yz

Is this solution Helpfull? Yes (3) | No (4)  

deepak  (9 years ago)

b)z but not y

Is this solution Helpfull? Yes (1) | No   

HARISHANKAR YADAV  (9 years ago)

as x=yz
((x-y)^n-(y-z)^n+z^n)/???
givn x=yz
nw ((yz-y)^n-(y-z)^n+z^n)/??
z^n((y-y/z)^n-(y/z-1)^n+1)/??
i think divisible by Z nt y

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

Sonali Ghadai  (9 years ago)

[x^n(+-)y^n-(y^n(+-)z^n)+z^n]-(.......)within this bracket whatever terms are there all contain both y and z,so this part divisible by y or z or yz.now the part within square bracket depend upon whether the value of n is even or odd(+ or _ depend upon this),if n is even then 1st part become
x^n+y^n-y^n-z^n+z^n=x^n=(yz)^n thus divisible by yor z or yz.

now,if n is odd,
x^n-y^n-y^n+z^n+z^n.now it is divisible by what?

upon this 1st part only ans is depended,because 2nd part always divisible by y or z or yz.

Is this solution Helpfull? Yes | No (6)  

LOKESH RANJAN  (9 years ago)

1,as put x=yz in equ^n we get it is always divisible by yz

Is this solution Helpfull? Yes | No (2)  

Anuja Singh  (9 years ago)

if n is odd then (yz^n............-y^n)-(y^n.........-z^n)+z^n
yz^n.....-2y^n......+2z^n then how will it be divided by yz??
can any bodu explain.........

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Ehtesham  (9 years ago)

take binomial expansion of (x-y)^n-(y-z)^n+z^n
then you will see divisibility depends on the nature of "n" .
ie . if n-even
then the whole expression will be divisible y;
and if n-odd
then there you can not reach to the result whether it gets divisible or not.

so in my opinion question is "aadha-adhura"

Is this solution Helpfull? Yes | No (1)  

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