Elitmus
Exam
Numerical Ability
Number System
(x-y)^n-(y-z)^n+z^n always divisible by what? given x=yz.
options:
1)yz
nt rember other options.
Read Solution (Total 10)
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- 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 - 10 years agoHelpfull: Yes(9) No(16)
- option
a)yz
b)z but not y
answer is b) z but not y - 10 years agoHelpfull: Yes(5) No(6)
- @ rakesh can u solve this???
- 10 years agoHelpfull: Yes(3) No(3)
- 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 - 10 years agoHelpfull: Yes(3) No(4)
- b)z but not y
- 10 years agoHelpfull: Yes(1) No(0)
- 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 - 10 years agoHelpfull: Yes(1) No(1)
- [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. - 10 years agoHelpfull: Yes(0) No(6)
- 1,as put x=yz in equ^n we get it is always divisible by yz
- 10 years agoHelpfull: Yes(0) No(2)
- 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......... - 10 years agoHelpfull: Yes(0) No(0)
- 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" - 9 years agoHelpfull: Yes(0) No(1)
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