EgmatQuantExpert wrote:

N is a positive two-digit number, whose tens digit is 9. If the units digit of \(N^p\) and \(N^q\) are 6 and 4 respectively. Find N.

(1) P is a positive even integer.

(2) Q is a positive odd integer.

\(\left\langle N \right\rangle = {\rm{units}}\,\,{\rm{digit}}\,\,{\rm{of}}\,\,N\,\,\,\)

\(\left. \begin{gathered}

N\,\, = \,\,\underline 9 \,\underline a \,\, \hfill \\

\left\langle {{N^p}} \right\rangle = 6\,\,\, \hfill \\

\left\langle {{N^q}} \right\rangle = 4 \hfill \\

\end{gathered} \right\}\,\,\,\,\,? = N\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\boxed{\,\,? = a\,\,}\)

\(\left( 1 \right)\,\,\,p \ge 2\,\,\,{\rm{even}}\,\,\,\left\{ \matrix{

\,\,\left\langle {{{92}^4}} \right\rangle = 6\,\,\,\,{\rm{and}}\,\,\,\,\,\left\langle {{{92}^2}} \right\rangle = 4\,\,\,\,\,\, \Rightarrow \,\,\,\,\,a = 2\,\,\,\,{\rm{viable}}\,\, \hfill \cr

\,\,\left\langle {{{94}^2}} \right\rangle = 6\,\,\,\,{\rm{and}}\,\,\,\,\,\left\langle {{{94}^1}} \right\rangle = 4\,\,\,\,\,\, \Rightarrow \,\,\,\,\,a = 4\,\,\,{\rm{viable}}\,\, \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{INSUFF}}.\,\)

\(\left( 2 \right)\,\,\,\left\{ \begin{gathered}

\,q \geqslant 1\,\,\,{\text{odd}} \hfill \\

\,\left\langle {{N^q}} \right\rangle = 4 \hfill \\

\end{gathered} \right.\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{inspection}}} \,\,\,\,\,a = 4\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\text{SUFF}}{\text{.}}\)

Obs.: inspection means: if

a = 0,1,2,3,5,6,7,8 or 9, when

a is put to an odd positive power, we don´t get units digit equal to 4, as needed.

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

Regards,

Fabio.

_________________

Fabio Skilnik :: https://GMATH.net (Math for the GMAT) or GMATH.com.br (Portuguese version)

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