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If (n+2)!= n!(an^2+bn+c), then abc=?

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Joined: 16 Aug 2015
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If (n+2)!= n!(an^2+bn+c), then abc=? [#permalink]

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12 Mar 2018, 01:04
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[GMAT math practice question]

If $$(n+2)!= n!(an^2+bn+c),$$ then $$abc=?$$

$$A. 2$$
$$B. 3$$
$$C. 4$$
$$D. 6$$
$$E. 8$$
[Reveal] Spoiler: OA

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"Only $79 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Manager Joined: 22 May 2017 Posts: 120 Re: If (n+2)!= n!(an^2+bn+c), then abc=? [#permalink] Show Tags 12 Mar 2018, 01:24 1 This post received KUDOS (n+2)(n+1)n!= n! (an2+bn+c) (n+2)(n+1) = (an2+bn+c) n2 + 3n +2 = (an2+bn+c) abc = 1*3*2 =6 Posted from my mobile device _________________ Kudos please if explanation helped ------------------------------------------------------------------------------------------------- Don't stop when you are tired , stop when you are DONE . Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 5267 GMAT 1: 800 Q59 V59 GPA: 3.82 Re: If (n+2)!= n!(an^2+bn+c), then abc=? [#permalink] Show Tags 14 Mar 2018, 00:24 => $$(n+2)! = (n+2)(n+1)n! = (n^2 + 3n + 2)n!$$ So, $$a = 1, b = 3$$, and $$c = 2$$. Thus, $$abc = 6.$$ Therefore, D is the answer. Answer: D _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 3 month Online Course"
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If (n+2)!= n!(an^2+bn+c), then abc=? [#permalink]

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14 Mar 2018, 09:08
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MathRevolution wrote:
[GMAT math practice question]

If $$(n+2)!= n!(an^2+bn+c),$$ then $$abc=?$$

$$A. 2$$
$$B. 3$$
$$C. 4$$
$$D. 6$$
$$E. 8$$

First recognize that (n + 2)! = (n + 2)(n + 1)(n)(n - 1)(n - 2)(n - 3)....(3)(2)(1)
= (n + 2)(n + 1)(n!)
= (n!)(n + 2)(n + 1)

The question tells us that (n + 2)! = n!(an² + bn + c)
So, we can write: (n!)(n + 2)(n + 1) = n!(an² + bn + c)
Divide both sides by n! to get: (n + 2)(n + 1) = an² + bn + c
Use FOIL to expand left side: n² + 3n + 2 = an² + bn + c
In other words: 1n² + 3n + 2 = an² + bn + c

So, a = 1, b = 3 and c = 2
This means abc = (1)(3)(2) = 6

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Re: If (n+2)!= n!(an^2+bn+c), then abc=? [#permalink]

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16 Mar 2018, 23:56
MathRevolution wrote:
[GMAT math practice question]

If $$(n+2)!= n!(an^2+bn+c),$$ then $$abc=?$$

$$A. 2$$
$$B. 3$$
$$C. 4$$
$$D. 6$$
$$E. 8$$

It can be written as:

$$\frac{(n + 2) (n + 1) (n)!}{n!}$$ $$=$$ $$(an^2+bn+c)$$

$$(n + 2) (n + 1)$$= $$(an^2+bn+c)$$

$$n^2 + 3n + 2 =$$ $$(an^2+bn+c)$$

Therefore, $$a = 1, b = 3, c = 2$$

(D)
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Re: If (n+2)!= n!(an^2+bn+c), then abc=?   [#permalink] 16 Mar 2018, 23:56
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