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4^N concept

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by arora007 » Tue Jul 06, 2010 6:25 am
If N is a positive integer, what is the last digit of 4^N ?

1. N^2 is divisible by 4
2. N + 2 is divisible by 6

[spoiler] it would not be tough to get that answer is D, but the take away is 4^N where N is even is alway gonna end in 6[/spoiler]

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by jeremy8 » Tue Jul 06, 2010 6:42 am
Establish a list of 4^N to see the cycle: 4,16,64,256,1024...
So when N is even, the last digit is always 6 and when N is odd, the last digit is 4.
We can rephrase as: Is N even or odd?

1) N^2 is divisible by 4 means N has at least one factor of 2, so Sufficient.
2) N+2 is divisible by 6 means that N is a multiple of 6, minus 2.
Since every multiple of 6 is even and since an even number -2 is always even as well, 2) is Sufficient.
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by arora007 » Tue Jul 06, 2010 7:00 am
yeah...We can rephrase as: Is N even or odd?

i.e.
4^N where N is even is alway gonna end in 6
4^N where N is odd is alway gonna end in 4
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