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

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Math Expert
Joined: 02 Sep 2009
Posts: 43378
f(x)=(4x+1)(4x–3) for all positive integers x. Which of the following [#permalink]

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21 Dec 2017, 19:25
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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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22 Dec 2017, 06:25
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.
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f(x)=(4x+1)(4x–3) for all positive integers x. Which of the following [#permalink]

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22 Dec 2017, 10:38
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

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

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, 10:38
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