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

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N is a two-digit number, whose tens digit is 9. If the units digit of  [#permalink]

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Take a stab at the following GMAT like question on Units Digit. The official solution will be posted soon.

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.


Answer Options
    A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
    B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
    C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
    D. EACH statement ALONE is sufficient to answer the question asked.
    E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.


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Originally posted by EgmatQuantExpert on 16 Dec 2016, 03:01.
Last edited by EgmatQuantExpert on 16 Dec 2016, 04:32, edited 5 times in total.
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N is a two-digit number, whose tens digit is 9. If the units digit of  [#permalink]

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New post Updated on: 25 Dec 2016, 06:49
Steps 1 & 2: Understand Question and Draw Inferences
Given: The tens digit of N is 9 and the unit’s digit of Np and Nq are 6 and 4 respectively.

To find: The value of N

Let us consider \(N = AB\), where A and B are single-digit numbers and the value of A is 9

Therefore, \(N^p = (AB)^p\)

To find the units digit, we need to focus on the last digit of the number, which is B in the above case. Hence, we can write \(B^p= 6\)

The above expression will hold true in the following cases –

Case I - Value of B = 2, Value of p = 4m
Case II - Value of B = 4, Value of p = 2m
Case III - Value of B = 6, Value of p = any number
Case IV - Value of B = 8, Value of p = 4m

From the above cases, we can infer that finding the value of B on the basis of the value of p is possible only when p is an odd number because –

If p is a multiple of 4, B can take values - 2,4,6 or 8
If p is a multiple of 2 but not a multiple of 4 then B can be either 4 or 6
If p is an odd number, B can take only one value - 6

Using the same logic, we can write \(N^q\) as \((AB)6^q\)  and \(B^q = 4\).
The above expression will hold true in the following cases –

Case I - Value of B = 2, Value of q = 4m+2
Case II - Value of B = 4, Value of q = 2m+1
Case III - Value of B = 8, Value of q = 4m+2

From the above cases, we can infer that finding the value of B on the basis of the value of q is only possible when q is an odd number because –

If q is an even number, B can be 2 or 8
If q is an odd number, B has to be 4


Step 3: Analyze Statement 1 independently
Statement 1 says “P is a positive even integer”.

From the inferences that we have drawn, we know if P is an even number, the units digit of N can be 2, 4, or 8

Since Statement 1 leads us to 4 possible values of B, it is clearly not sufficient to arrive at a unique answer.

Step 4: Analyze Statement 2 independently

Statement 2 says that Q is a positive odd integer.

From the inferences that we have drawn, we know if Q is an odd number, the units digit of N has to be 4

Thus N = 94.

Statement 2 is sufficient to get us a unique answer.

Step 5: Analyze Both Statements Together (if needed)

Since we’ve already got a unique answer in Step 4, this step is not required

Hence Correct Answer: Option B
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Originally posted by EgmatQuantExpert on 16 Dec 2016, 03:02.
Last edited by EgmatQuantExpert on 25 Dec 2016, 06:49, edited 1 time in total.
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Re: N is a two-digit number, whose tens digit is 9. If the units digit of  [#permalink]

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New post 16 Dec 2016, 05:00
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.
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Re: N is a two-digit number, whose tens digit is 9. If the units digit of  [#permalink]

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New post 16 Dec 2016, 05:21
koenh wrote:
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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Re: N is a two-digit number, whose tens digit is 9. If the units digit of  [#permalink]

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New post 25 Dec 2016, 06:49
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N is a two-digit number, whose tens digit is 9. If the units digit of  [#permalink]

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New post 28 Oct 2018, 14:08
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.
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Re: N is a two-digit number, whose tens digit is 9. If the units digit of  [#permalink]

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Re: N is a two-digit number, whose tens digit is 9. If the units digit of   [#permalink] 28 Jan 2020, 14:29
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