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

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Math Revolution GMAT Instructor
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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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Question Stats:

88% (01:17) correct 13% (01:50) wrong based on 64 sessions

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[GMAT math practice question]

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

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

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"Only $79 for 1 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: 119 Re: If (n+2)!= n!(an^2+bn+c), then abc=? [#permalink] ### Show Tags 12 Mar 2018, 01:24 2 (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: 7618 GMAT 1: 760 Q51 V42 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 1 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
Top Contributor
1
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]

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09 Mar 2019, 19:59
Hi chetan2u,

I was wondering could you please explain where I am making a mistake in my solution? I am getting different answers.

Let's say n=1

6=1(1a+1b+c). If we were to pick a=b=1 and c=4, abc=6 BUT if we were to pick a=b=c=2, we get 8 instead.

Can we not pick smart numbers for this question specifically?
Math Expert
Joined: 02 Aug 2009
Posts: 7763
Re: If (n+2)!= n!(an^2+bn+c), then abc=?  [#permalink]

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09 Mar 2019, 20:39
JS1290 wrote:
Hi chetan2u,

I was wondering could you please explain where I am making a mistake in my solution? I am getting different answers.

Let's say n=1

6=1(1a+1b+c). If we were to pick a=b=1 and c=4, abc=6 BUT if we were to pick a=b=c=2, we get 8 instead.

Can we not pick smart numbers for this question specifically?

If you want to solve by plugging numbers here, you should look at for possible answer but then test these numbers in other values of n..
for example take n as 2 now..
$$(2+2)!=2!(a2^2+2b+c).......24=2(4a+2b+c)..........12=4a+2b+c$$
the values of (a, b, c) when n=1 were (1, 1, 4) or (1,4,1) or (4,1,1) or (2,2,2) or (1,2,3) or (1,3,2) and so on..
Only (1,3,2) satisfies the equation when n=2.

So, the method of plugging would become a bit more cumbersome, although it will give you a correc answer after few steps
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Re: If (n+2)!= n!(an^2+bn+c), then abc=?   [#permalink] 09 Mar 2019, 20:39
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