N is a two-digit number, whose tens digit is 9. If the units

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AAPL wrote:e-GMAT

N is a two-digit positive 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 :: GMATH method creator ( Math for the GMAT)
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by Jay@ManhattanReview » Sun Oct 28, 2018 10:27 pm

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AAPL wrote:e-GMAT

N is a 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.

OA B.
Say N = 9X, where X is the unit digit of N.

Given:

1. Units digit of N^p is 6 and
2. Units digit of N^q is 4

Note that unit digits of N^p and N^q will be determined by the units' digit of N, i.e the value of X.

Since the unit digits of N^p and N^q are 6 and 4 (EVEN), X must be even.

Let's list down the unit digit power cycle of 2, 4, 6, and 8.

2: (2, 4, 8, 6), (2, 4, 8, 6), (2, 4, 8, 6), ... => We see that 4 and 6 are there, thus, X can be 2.
4: (4, 6), (4, 6),(4, 6), ... => We see that 4 and 6 are there, thus, X can be 4.
6: 6, 6, 6, ... => We see that 4 is not there, thus, X cannot be 6.
8: (8, 4, 2, 6), (8, 4, 2, 6), (8, 4, 2, 6), ..... => We see that 4 and 6 are there, thus, X can be 8.

Thus, X is one among 2, 4 and 8.

Question rephrased: Which among 2, 4 and 8 is X?

Let's take each statement one by one.

1) P is a positive even integer.

=> P = 2, 4, 6, ...

Since the unit digit of N^p is 6 (Note the occurrence of 8 in blue at even places), X can be 2, 4 or 8. No unique value of X. Insufficient.

2) Q is a positive odd integer.

X can only be 4 since N^1, N^3 or N^5 will render 4 as its unit digit if N = 94. Sufficient.

The correct answer: B

Hope this helps!

-Jay
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