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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 \(abc=?\)

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

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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] 16 Mar 2018, 23:56
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