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If x is a positive integer and f(x)=x–x2–x3+x4+x5–x6–x7+x8, then is f(x) divisible by 96?


(1) x–x2–x3 is divisible by 32

(2) x has no prime factors other than 2

96=32*3
1) insufficient
2) 96 has 3 in prime factors
so answer is not divisible by 96
sufficient
Ans B
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Quote:
If x is a positive integer and \(f(x) = x – x^2 – x^3 + x^4 + x^5 – x^6 – x^7 + x^8\), then is f(x) divisible by 96?


(1) \(x – x^2 – x^3\) is divisible by 32

(2) x has no prime factors other than 2

\(f(x) = x – x^2 – x^3 + x^4 + x^5 – x^6 – x^7 + x^8\)
=\(x+x^5-x^2-x^6-x^3-x^7+x^4+x^8\)
=\(x(1+x^4)-x^2(1+x^4)-x^3(1+x^4)+x^4(1+x^4)\)
=\((1+x^4)(x-x^2-x^3+x^4)\)=\((1+x^4)(x(1-x)-x^3(1-x)\)
=\((1+x^4)(1-x)(x-x^3)\)
=\((1+x^4)(1-x)x(1-x)(1+x)\)
=\(x(x-1)^2(x+1)(1+x^4)\)
Now, (x-1)x(x+1) is product of three consecutive integers, so surely a multiple of 6.
Thus we are looking at if the term is a multiple of 96/6=16.

Quote:
If you have a problem in factorizing, use remainders when divided by 2 and 3 as \(96=2^5*3\)..
1) divisibility by 2 :- there are 8 terms in x, so surely a multiple of 2
2) Divisibility by 3 :-
a) If x is divisible by 3, entire term is divisible by 3.
b) If x leaves a remainder 1, then \(f(x) = x – x^2 – x^3 + x^4 + x^5 – x^6 – x^7 + x^8\) leaves a remainder \( 1 – 1^2 – 1^3 + 1^4 + 1^5 – 1^6 – 1^7 + 1^8=0\)
c) If x leaves a remainder 2, then \(f(x) = x – x^2 – x^3 + x^4 + x^5 – x^6 – x^7 + x^8\) leaves a remainder \( 2 – 2^2 – 2^3 + 2^4 + 2^5 – 2^6 – 2^7 + 2^8=2-1-2+1+2-1-2+1=0\)
So, in all cases surely divisible by 3.

From both of the methods above, we know that f(x) is divisible by 6 and we have to check whether \(x(x-1)^2(x+1)(1+x^4)\) is divisible by 32.
Some additional information
1) If x is odd, each of \((x-1)^2,(x+1),(1+x^4)\) will be divisible by 2, and atleast one of x-1 or x+1 will be surely divisible by 4. So the term will be divisible by 2*2*4*2=32 for sure. Hence we will be able to answer YES
2)If x is even, each of \((x-1)^2,(x+1),(1+x^4)\) will be odd, and divisibility by 32 will depend on just x, that is
a) If x is a multiple of 32..YES, otherwise NO

So requirement..
1) Is x odd? then yes
2) If even, is x a multiple of 32.


(1) \(x – x^2 – x^3\) is divisible by 32
\(x – x^2 – x^3=x(1-x-x^2)\)
If x is even, 1-x-x^2 is odd, and so x is divisible by 32.
If x is odd, 1-x-x^2 is also odd, so discard as x(1-x-x^2) will not be divisible by 32.
So x is multiple of 32.
Suff

(2) x has no prime factors other than 2
We do not know whether x is a multiple of 32.
Insuff

A
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f(x) = x – x^2 – x^3 + x^4 + x^5 – x^6– x^7+ x^8
=(x – x^2 – x^3)+x^4(x – x^2 – x^3)+x^4(x^4+1)
=(x – x^2 – x^3)(x^4+1)+x^4(x^4+1)
=(x^4+1)(x – x^2 – x^3+x^4)
=(x^4+1)(-x(x-1)+x^3(x-1))
=(x^4+1)(x-1)(x^3-x)
= (x^4+1)(x-1)(x^2-1)x
=[(x^4+1)(x-1)(x-1)(x+1)] x
= A * B [ where A = (x^4+1)(x-1)(x-1)(x+1) and B= x]

as , x has no prime factor other than 2, then x=2^n ,where n= 1,2,3,4,5

Therefore, A is always an odd number and divisible by 3. so if f(x) has to be divisible by 96(3*32) , then x has to be divisible by 32.
Hence x will be multiplier of 32 e.g 32,64,96,

Now as x is divisible by 32, x-x^2-x^3 is also divisible by 32. therefore this condition is redundant.

Hence correct option would be : B
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