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

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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, $$abc = 6.$$

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