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

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

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

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

\(A. 2\)
\(B. 3\)
\(C. 4\)
\(D. 6\)
\(E. 8\)

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

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New post 12 Mar 2018, 01:24
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(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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Re: If (n+2)!= n!(an^2+bn+c), then abc=?  [#permalink]

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

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New post 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

Answer: D

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

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

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New post 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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