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# x is an integer. If [m]x>9[/m], and [m]x^3+3x^2+2x[/m] is no

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x is an integer. If [m]x>9[/m], and [m]x^3+3x^2+2x[/m] is no [#permalink]  13 Aug 2019, 06:03
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100% (00:37) correct 0% (00:00) wrong based on 2 sessions
x is an integer. If $$x>9$$, and $$x^3+3x^2+2x$$ is not divisible by 4, then which of the following MUST be divisible by 4?

A) $$x-3$$
B) $$x-4$$
C) $$x-5$$
D) $$x-6$$
E) $$x-7$$
[Reveal] Spoiler: OA

_________________

Brent Hanneson – Creator of greenlighttestprep.com

GRE Instructor
Joined: 10 Apr 2015
Posts: 2290
Followers: 71

Kudos [?]: 2165 [1] , given: 26

Re: x is an integer. If [m]x>9[/m], and [m]x^3+3x^2+2x[/m] is no [#permalink]  13 Aug 2019, 06:15
1
KUDOS
Expert's post
GreenlightTestPrep wrote:
x is an integer. If $$x>9$$, and $$x^3+3x^2+2x$$ is not divisible by 4, then which of the following MUST be divisible by 4?
A) $$x-3$$
B) $$x-4$$
C) $$x-5$$
D) $$x-6$$
E) $$x-7$$

Solution #1: Algebraic

Key concept: When we list consecutive integers, every 4th value is a multiple of 4
{...-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,...}
So, among any 4 consecutive integers, there will be ONE value that is divisible by 4.

-----------ONTO THE QUESTION------------------------

Take $$x^3+3x^2+2x$$
Factor out the x to get: $$x(x^2+3x+2)$$
Factor the quadratic to get: $$x(x+1)(x+2)$$
Notice that x, (x+1), and (x+2) are CONSECUTIVE integers.
Since we're told that $$x^3+3x^2+2x$$ is not divisible by 4, we know that none of the 3 values, x, (x+1), or (x+2), are divisible by 4.
So, among the FOUR consecutive integers (x-1), (x), (x+1), and (x+2), we know that x-1 must be divisible by 4 (since none of the other 3 values are divisible by 4)
In fact, if we list more consecutive integers in this form, we'll see that every FOURTH number will be divisible by 4.

We get: {...., (x-9), (x-8), (x-7), (x-6), (x-5), (x-4), (x-3), (x-2), (x-1), (x), (x+1), (x+2), (x+3),...}

So, we can see that (x-5) must be divisible by 4

Cheers,
Brent
_________________

Brent Hanneson – Creator of greenlighttestprep.com

GRE Instructor
Joined: 10 Apr 2015
Posts: 2290
Followers: 71

Kudos [?]: 2165 [0], given: 26

Re: x is an integer. If [m]x>9[/m], and [m]x^3+3x^2+2x[/m] is no [#permalink]  13 Aug 2019, 06:23
Expert's post
GreenlightTestPrep wrote:
x is an integer. If $$x>9$$, and $$x^3+3x^2+2x$$ is not divisible by 4, then which of the following MUST be divisible by 4?
A) $$x-3$$
B) $$x-4$$
C) $$x-5$$
D) $$x-6$$
E) $$x-7$$

Solution #2: Testing values

Take $$x^3+3x^2+2x$$
Factor out the x to get: $$x(x^2+3x+2)$$
Factor the quadratic to get: $$x(x+1)(x+2)$$
Notice that x, (x+1), and (x+2) are CONSECUTIVE integers.

So, let's find 3 consecutive numbers that meet the condition that $$x(x+1)(x+2)$$ is not divisible by 4
For this to happen, none of the 3 values can be divisible by 4.

Well, 13, 14 and 15 meet this condition.
In this case, x = 13, x+1 = 14 and x+2 = 15

We can now check the answer choices by plugging x = 13 into each answer choice.

A) x - 3 = 13 - 3 = 10. 10 is NOT divisible by 4. Eliminate A.

B) x - 4 = 13 - 4 = 9. 9 is NOT divisible by 4. Eliminate B.

C) x - 5 = 13 - 5 = 8. 8 IS divisible by 4!!

Cheers,
Brent
_________________

Brent Hanneson – Creator of greenlighttestprep.com

Intern
Joined: 24 Jul 2019
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Kudos [?]: 2 [2] , given: 1

Re: x is an integer. If [m]x>9[/m], and [m]x^3+3x^2+2x[/m] is no [#permalink]  14 Aug 2019, 05:29
2
KUDOS
I have a slightly different way of solving the question.

I started by plugging in possible numbers for x in x^3+3x^2+2x. I started with 10 and 11, found out that the answers are divisible by 4, then tried 17 which intuitively felt like there is a higher possibility of it not being able to be divided by 4.
Indeed, 17^3+3(17)^2+2(17)= 5814, which is not divisible by 4.

I then looked at the answer options for the next step- since the answer options are of numbers which are smaller than x, I had to think of a number that is smaller than 17 and is a multiple of 4. 12 is smaller than 17 and is divisible by 4 and 17-12=5
Therefore, x-5 MUST be divisible by 4 and (C) is the correct answer.
GRE Instructor
Joined: 10 Apr 2015
Posts: 2290
Followers: 71

Kudos [?]: 2165 [0], given: 26

Re: x is an integer. If [m]x>9[/m], and [m]x^3+3x^2+2x[/m] is no [#permalink]  14 Aug 2019, 05:43
Expert's post
gwent wrote:
I have a slightly different way of solving the question.

I started by plugging in possible numbers for x in x^3+3x^2+2x. I started with 10 and 11, found out that the answers are divisible by 4, then tried 17 which intuitively felt like there is a higher possibility of it not being able to be divided by 4.
Indeed, 17^3+3(17)^2+2(17)= 5814, which is not divisible by 4.

I then looked at the answer options for the next step- since the answer options are of numbers which are smaller than x, I had to think of a number that is smaller than 17 and is a multiple of 4. 12 is smaller than 17 and is divisible by 4 and 17-12=5
Therefore, x-5 MUST be divisible by 4 and (C) is the correct answer.

Great reasoning!

Cheers,
Brent
_________________

Brent Hanneson – Creator of greenlighttestprep.com

Re: x is an integer. If [m]x>9[/m], and [m]x^3+3x^2+2x[/m] is no   [#permalink] 14 Aug 2019, 05:43
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