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

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Question Stats: 86% (01:17) correct 14% (01:36) wrong based on 69 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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Re: If (n+2)!= n!(an^2+bn+c), then abc=?  [#permalink]

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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

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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8009
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: If (n+2)!= n!(an^2+bn+c), then abc=?  [#permalink]

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=>

$$(n+2)! = (n+2)(n+1)n! = (n^2 + 3n + 2)n!$$
So,
$$a = 1, b = 3$$, and $$c = 2$$.
Thus, $$abc = 6.$$

Therefore, D is the answer.

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

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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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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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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?
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Joined: 02 Aug 2009
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Re: If (n+2)!= n!(an^2+bn+c), then abc=?  [#permalink]

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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
_________________ Re: If (n+2)!= n!(an^2+bn+c), then abc=?   [#permalink] 09 Mar 2019, 20:39
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