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#7 from Number Properties - II [#permalink]
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.

Re: #7 from Number Properties - II [#permalink]
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?

gmatclubot

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