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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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14% (00:04) wrong based on 7 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) \(x7\)
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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
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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
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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
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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
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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.



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