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What is the value of integer n ? (1) n(n + 1) = 20 (2) 6!/

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by GMATinsight » Sat Sep 08, 2018 4:28 am

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What is the value of integer n ?

(1) n(n + 1) = 20
(2) 6!/(n!)(6-n)! = 15

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What is the value of integer n ? (1) n(n + 1) = 20 (2) 6!/

by fskilnik@GMATH » Sat Sep 08, 2018 3:11 pm
GMATinsight wrote:What is the value of integer n ?

(1) n(n + 1) = 20
(2) 6!/[(n!)(6-n)!] = 15

Source: www.GMATinsight.com
Obs.: in GMAT`s universe, factorials are defined ONLY for nonnegative integer values.
Therefore (when considering) statement (2) we MUST implicitly ASSUME that n is equal to 0,1,2,3,4,5 or 6.

\[\left( 1 \right)\,\,\,20 = \left\{ \begin{gathered}
\,\,4 \cdot 5\,\,\, \Rightarrow \,\,\,n\, = 4\,\, \hfill \\
\left( { - 5} \right) \cdot \left( { - 4} \right)\,\,\, \Rightarrow \,\,\,n = \, - 5\,\,\, \hfill \\
\end{gathered} \right.\,\,\,\, \Rightarrow \,\,\,\,\,\,{\text{INSUF}}.\,\,\]

Important: we found TWO distinct real roots for the SECOND-degree equation given in statement (1).
Conclusion: there are NO other possibilities for n (even when nonintegers are considered, by the way!)

\[\left( 2 \right)\,\,15 = \frac{{6!}}{{n!\,\,\left( {6 - n} \right)!}} = C\left( {6,n} \right)\,\,\,\, \Rightarrow \,\,\,n = 2\,\,\,{\text{or}}\,\,\,n = 4\,\,\,\, \Rightarrow \,\,\,\,{\text{INSUF}}.\,\,\,\]

\[\left( {1 + 2} \right)\,\,\,\,\,\,\,n = 4\,\,\,\, \Rightarrow \,\,\,\,{\text{SUF}}.\]

The right answer is therefore [spoiler]__(C)__[/spoiler].


This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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