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f(x)=(4x+1)(4x–3) for all positive integers x. Which of the following

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f(x)=(4x+1)(4x–3) for all positive integers x. Which of the following [#permalink]

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New post 21 Dec 2017, 20:25
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A
B
C
D
E

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\(f(x)=(4x+1)(4x–3)\) for all positive integers x. Which of the following cannot be a factor of f(x)?

I. 7
II. 16
III. 25

A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III
[Reveal] Spoiler: OA

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f(x)=(4x+1)(4x–3) for all positive integers x. Which of the following [#permalink]

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New post 22 Dec 2017, 07:25
Expert's post
As it can often be hard to understand the rules underlying divisibility, we'll try a few numbers to see the pattern.
This is an Alternative approach.

Starting with x = 1, we get
5*1. Not divisible by 7,16 or 25.
9*5. Not divisible by 7,16 or 25.
13*9. Not divisible by 7,16 or 25.
17*13. Not divisible by 7,16 or 25.
21*17. Divisbile by 7!
25*21. Divisbile by 25!
So all we need to know is if f(x) can be divisible by 16.
Trying a few more options, we have
29*25
33*29
37*33
We've already tried quite a few numbers without finding one that is divisible by 16 so we should feel comfortable marking (B).
In fact, since neither of our numbers will ever be divisible by 4 (as they are 4x+1 and 4x-3), their product cannot be divisible by 16.
(B) is our answer.
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f(x)=(4x+1)(4x–3) for all positive integers x. Which of the following [#permalink]

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New post 22 Dec 2017, 11:38
Bunuel wrote:
\(f(x)=(4x+1)(4x–3)\) for all positive integers x. Which of the following cannot be a factor of f(x)?

I. 7
II. 16
III. 25

A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III


Working through numbers has already been explained so here's an alternate approach

\(4x+1\) & \(4x-3\) are ODD as \(x\) is an integer. Product of Odd numbers will not be divisible by Even number. So \(16\) cannot be a factor of \(f(x)\)

So options A & D are out. Now I is present in options C & E, if we can eliminate I then the answer will be B.

if \(7\) is a factor of \(f(x)\), then either \(4x+1\) or \(4x-3\) has to be a multiple of \(7\)

so if, \(4x-1=7k => x=\frac{7k+1}{4}\), where \(x\) has to be an integer. clearly if \(k=1\) then \(x=2\), an integer

or \(4x-3=7k=> x=\frac{7k+3}{4}\), if \(k=3\) then \(x=6\), an integer. So we can conclude \(7\) can be a factor of \(f(x)\)

So we have eliminated Option A, C, D & E

Answer: Option B

---------------------------------------------------------

just to test \(25\) as a factor of \(f(x)\):

Now if \(4x+1=25 => x=6\) an integer so \(25\) can be a factor of \(4x+1\). Similarly if \(4x-3=25=>x=7\). hence \(25\) can also be a factor of \(4x-3\). so we know \(25\) can be a factor of \(f(x)\)
f(x)=(4x+1)(4x–3) for all positive integers x. Which of the following   [#permalink] 22 Dec 2017, 11:38
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