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koenh
Let N = ab and a = 9. Now N = 9b. We need to find b to find N.

Given: \(N^p = 6\) and \(N^q = 4\)

So N needs to have unit digits 6 and 4 if raised to certain powers.
This gives three options:
b = 2, 4 or 8.
For example:
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 6

1) p = even. Out of the three options, 2^4 (even) = 6 and 4^2 (even) = 6 and 8^4 (even) = 6. So insufficient.

2) q = odd. Out of the three options, there is only one b that produces unit digit 4 when raised to an odd power:
2 raised to an odd power alternatives between 2 and 8.
8 raised to an odd power alternatives between 8 and 2.
4 raised to an odd power is always 4.

So statement 2 is sufficient and answer is B.

Hi,

Good analysis. You've solved the question correctly and you've taken into account all possible cases while solving it. The best part of your solution is that you've done a thorough Question Statement analysis before moving on to analyze the Statements (1) and (2) independently. Doing so reduces the chances of making an error while solving DS questions.

Good job :)

Regards,
Saquib
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The detailed solution has been posted. Go through it and let us know if you have any queries.
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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.
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