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#7 from Number Properties - II

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#7 from Number Properties - II [#permalink] New post 26 Jul 2010, 23:16
Hello! I'm not sure what the proper tag for this question is since it's in the "Number Properties - II" bin rather than one of the official exams. I think I have found a problem with the OA.

Here's the question:


If n is a positive integer, what is the last digit of 4n?

1.) n^2 is divisible by 4
2.) n+2 is divisible by 6




The OA is D, but I have a bit of a problem with Statement 1.

The explanation states that the last digit of 4 raised to any positive even integer is 6. While I agree with this statement, I believe that this overlooks the possibility that n=2.
If n=2:

1) n^2 is divisible by 4
(2^2) / 4 = 4/4 = 1

The last digit of 4n is therefore 8 (4 * 2 = 08).

Let's try n=4 to double check that the answer isn't always 8.
If n=4:

1) n^2 is divisible by 4
16/4 = 4

The last digit of 4n is therefore 6 (4*4 = 16)


I'm sorry for the lack of formatting, but I wanted to type this out really quickly before I go to bed. I'll come back and change it when I'm more awake.
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Re: #7 from Number Properties - II [#permalink] New post 27 Jul 2010, 01:57
Are you sure the answer for this question is D?
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Re: #7 from Number Properties - II [#permalink] New post 27 Jul 2010, 09:07
Hi,
by 4n, shouldn't we understand the 2 digit number 4n ? By doing so, D seems to be the answer.
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Re: #7 from Number Properties - II [#permalink] New post 27 Jul 2010, 09:26
Geronimo wrote:
Hi,
by 4n, shouldn't we understand the 2 digit number 4n ? By doing so, D seems to be the answer.


Nope. If we understand it as the 2-digit number 4n (a.k.a. (40 + n)), then n could be 2, 4, 6, or 8, as all of these numbers, when squared, are divisible by 4. The answer wouldn't be D if this were the case.

I think it's an error in the answer.

Anyone else?
Re: #7 from Number Properties - II   [#permalink] 27 Jul 2010, 09:26
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