Re: Is the integer n odd? (1) n is divisible by 3 (2) 2n is divisible by
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15 Nov 2008, 12:29
Statement #1 is not sufficient for reasons mentioned by icandy..so A and D are out as choices
Statement #2 gives an interesting data point. It says 2n has exactly 2xfactors compared to n
Let's take n = 6, then factors are 1, 2, 3 and 6 itself for a total count of 4
Now 2n = 12, so factors are 1, 2, 3, 4, 6, 12 for a total of 6.
Another sample, say n = 8, then factors are 1, 2, 4, 8 (count 4)
2n = 16, then factors are 1, 2, 4, 8, 16 (count 5).
Hmmm ok, doesn't look like even numbers satisfy this. Let's try with a couple of odd number substitutions
Say, n = 5, then factors are 1, 5(count = 2)
2n = 10, factors are 1, 2, 5, 10 (count = 4)...looking good so far
Say n = 9, then factors are 1, 3, 9(count = 3)
2n = 18, factors are 1, 2, 3, 6, 9, 18(count = 6). hmmm good are we ready to call n odd.
Let's hold on and do a prime number test
Say n = 3, then factors are 1, 3(count = 2)
2n = 6, then factors are 1, 2, 3, 6(count = 4). Good let's do the last prime # try
Say n = 2, then factors are 1, 2(count = 2)
2n = 4, then factors are 1, 2, 4 (count = 3)
Seems like #2 is sufficient to answer the question whether n is odd. So pick is B
It'd be a better question, if it was asked whether n is a prime number? That'd have made it a little bit more tricky, by extending us to use both #1 and #2. In which case I think I'd go with C.