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18 May 2020, 01:58
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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
Are You Up For the Challenge: 700 Level Questions
18 May 2020, 16:23
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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^4 *(x – x^2 – x^3 + x^4)
= (1+x^4) * (x – x^2 – x^3 + x^4)
= (1+x^4) * (x-1)* {(x-1) * x * (x+1)}
Q. If f(x) divisible by 96=3*32 ?
(1) x * (1 – x – x^2) is divisible by 32
--> x = 32, 64, 96, etc.
x=32... f(x) = (1+32^4) * 31 * {31 * 32 * 33}
--> f(x) is divisible by 96.
x=64... f(x) = (1+64^4) * 64 * {63 * 64 * 65}
--> f(x) is divisible by 96.
et cetera...
SUFFICIENT
(2) x has no prime factors other than 2
x=2... f(x) = (1+2^4) * 2 * {1 * 2 * 3}
--> f(x) is NOT divisible by 96.
x=32... f(x) = (1+32^4) * 31 * {31 * 32 * 33}
--> f(x) is divisible by 96.
NOT SUFFICIENT
FINAL ANSWER IS (A)
Posted from my mobile device
= (x – x^2 – x^3 + x^4) + (x^5 – x^6 – x^7 + x^8)
= (x – x^2 – x^3 + x^4) + x^4 *(x – x^2 – x^3 + x^4)
= (1+x^4) * (x – x^2 – x^3 + x^4)
= (1+x^4) * (x-1)* {(x-1) * x * (x+1)}
Q. If f(x) divisible by 96=3*32 ?
(1) x * (1 – x – x^2) is divisible by 32
--> x = 32, 64, 96, etc.
x=32... f(x) = (1+32^4) * 31 * {31 * 32 * 33}
--> f(x) is divisible by 96.
x=64... f(x) = (1+64^4) * 64 * {63 * 64 * 65}
--> f(x) is divisible by 96.
et cetera...
SUFFICIENT
(2) x has no prime factors other than 2
x=2... f(x) = (1+2^4) * 2 * {1 * 2 * 3}
--> f(x) is NOT divisible by 96.
x=32... f(x) = (1+32^4) * 31 * {31 * 32 * 33}
--> f(x) is divisible by 96.
NOT SUFFICIENT
FINAL ANSWER IS (A)
Posted from my mobile device
General Discussion
18 May 2020, 04:08
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\(f(x) = x – x^2 – x^3 + x^4 + x^5 – x^6 – x^7 + x^8\)
\(f(x) = x(1 – x – x^2 + x^3 + x^4 – x^5 – x^6 + x^7)\)
\(f(x) = x[(1 – x – x^2) + x^3 + x^4(1 – x – x^2) + x^7]\)
\(f(x) = x[(1 – x – x^2) + x^4(1 – x – x^2)+x^3 + x^7]\)
\(f(x) = x[(x^4+1)(1 – x – x^2) + x^3(1 + x^4)]\)
\(f(x) = x[(x^4+1)(1 – x – x^2+x^3)]\)
\(f(x) = x[(x^4+1)(x^3 – x^2 – x++1)]\)
\(f(x) = x[(x^4+1)(x^2(x – 1) –1( x-1)]\)
\(f(x) = x[(x^4+1)(x^2 – 1) ( x-1)]\)
\(f(x) = x[(x^4+1)(x+1) ( x-1)^2]\)
f(x) is a multiple of \((x-1)*x*(x+1)\). Hence, it must be a multiple of 3.
Our question stem is whether f(x) is divisible by 32.
Statement 1-
\(x(1-x-x^2)\) is multiple of 32
\((1-x-x^2) = [1-(x)(x+1] \)
x(x+1) is always even{product of 2 consecutive numbers}; hence [1-(x)(x+1)] must be odd.
So x is a multiple of 32.
x is multiple of 32 and f(x) is multiple of x. So, f(x) must be a multiple of 32.
Sufficient.
Statement 2-
If x =2 f(x) is not a multiple of 32
If x=32. f(x) is a multiple of 32
Insufficient
\(f(x) = x(1 – x – x^2 + x^3 + x^4 – x^5 – x^6 + x^7)\)
\(f(x) = x[(1 – x – x^2) + x^3 + x^4(1 – x – x^2) + x^7]\)
\(f(x) = x[(1 – x – x^2) + x^4(1 – x – x^2)+x^3 + x^7]\)
\(f(x) = x[(x^4+1)(1 – x – x^2) + x^3(1 + x^4)]\)
\(f(x) = x[(x^4+1)(1 – x – x^2+x^3)]\)
\(f(x) = x[(x^4+1)(x^3 – x^2 – x++1)]\)
\(f(x) = x[(x^4+1)(x^2(x – 1) –1( x-1)]\)
\(f(x) = x[(x^4+1)(x^2 – 1) ( x-1)]\)
\(f(x) = x[(x^4+1)(x+1) ( x-1)^2]\)
f(x) is a multiple of \((x-1)*x*(x+1)\). Hence, it must be a multiple of 3.
Our question stem is whether f(x) is divisible by 32.
Statement 1-
\(x(1-x-x^2)\) is multiple of 32
\((1-x-x^2) = [1-(x)(x+1] \)
x(x+1) is always even{product of 2 consecutive numbers}; hence [1-(x)(x+1)] must be odd.
So x is a multiple of 32.
x is multiple of 32 and f(x) is multiple of x. So, f(x) must be a multiple of 32.
Sufficient.
Statement 2-
If x =2 f(x) is not a multiple of 32
If x=32. f(x) is a multiple of 32
Insufficient
