Aug 17 08:00 PM PDT  09:00 PM PDT Anyone who signs up for our free 7day trial automatically gets access to one free gre practice test. When you upgrade to a full Economist GRE Tutor plan, you will have a choice of plans that offer between three and five practice exams. Aug 17 02:00 PM EDT  04:00 PM EDT Our seminars give you the inside scoop on the exams, so you’ll understand everything you need to succeed. Saturday, August 17th at 2  4 PM ET Aug 19 08:00 AM PDT  09:00 AM PDT Join a 4day FREE online boot camp to kick off your GRE preparation and get you into your dream grad school in the upcoming deadlines.**Limited for the first 99 registrants. Aug 20 08:00 PM PDT  09:00 PM PDT The GRE is a standardized test that covers a broad range of quantitative and verbal topics and requires ample preparation time, so you need to be strategic about when to take it. Aug 23 08:00 PM PDT  10:00 PM PDT Get Magoosh's App for Free Questions, Free Video lessons, and more. Aug 25 09:00 PM PDT  10:00 PM PDT Target Test Prep is giving 10 lucky winners 4 months of FREE access to our toprated GRE Quant course. Our giveaway contest is open for 10 days only, so enter today for your chance to levelup your prep! Aug 26 08:00 PM PDT  09:00 PM PDT We have a growing team of trained online experts, but much of our online GRE tutoring is delivered by a true GRE expert, MyGuru's Director of Online Tutoring, Stefan. Aug 28 08:00 PM PDT  11:00 PM PDT Free Video Modules Within each of the 16 learning modules (Geometry, Statistics, Sentence Equivalence, etc.) that comprise the course, you'll find plenty of free videos to help you make an informed purchase.
Author 
Message 
TAGS:


GRE Instructor
Joined: 10 Apr 2015
Posts: 2290
Followers: 71
Kudos [?]:
2165
[0], given: 26

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
Question Stats:
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) \(x3\) B) \(x4\) C) \(x5\) D) \(x6\) E) \(x7\)
_________________
Brent Hanneson – Creator of greenlighttestprep.com Sign up for my free GRE Question of the Day emails




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
This post received KUDOS
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) \(x3\) B) \(x4\) C) \(x5\) D) \(x6\) E) \(x7\) 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 (x1), (x), (x+1), and (x+2), we know that x1 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: {...., (x9), (x8), (x7), (x6), (x5), (x4), (x3), (x2), (x1), (x), (x+1), (x+2), (x+3),...} So, we can see that (x5) must be divisible by 4 Answer: C Cheers, Brent
_________________
Brent Hanneson – Creator of greenlighttestprep.com Sign up for my free GRE Question of the Day emails



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
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) \(x3\) B) \(x4\) C) \(x5\) D) \(x6\) E) \(x7\) 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!! Answer: C Cheers, Brent
_________________
Brent Hanneson – Creator of greenlighttestprep.com Sign up for my free GRE Question of the Day emails



Intern
Joined: 24 Jul 2019
Posts: 1
Followers: 0
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
This post received 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 1712=5 Therefore, x5 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
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 1712=5 Therefore, x5 MUST be divisible by 4 and (C) is the correct answer. Great reasoning! Cheers, Brent
_________________
Brent Hanneson – Creator of greenlighttestprep.com Sign up for my free GRE Question of the Day emails




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





