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Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
x is an integer. If [m]x>9[/m], and [m]x^3+3x^2+2x[/m] is no
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13 Aug 2019, 06:03
13 Aug 2019, 06:03
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Question Stats:
76% (01:57) correct
23% (01:24) wrong based on 13 sessions
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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) \(x-7\)
A) \(x-3\)
B) \(x-4\)
C) \(x-5\)
D) \(x-6\)
E) \(x-7\)
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
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
13 Aug 2019, 06:15
1
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) \(x-7\)
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
Answer: C
Cheers,
Brent
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
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
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) \(x-3\)
B) \(x-4\)
C) \(x-5\)
D) \(x-6\)
E) \(x-7\)
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!!
Answer: C
Cheers,
Brent
