Just for fun consider the following in an expanded

[JohnHarris]

Just for fun consider the following in an expanded solution set that contains zero [you will only be tested on divisibility of positive numbers, see the link below].

Is the integer n odd?
1. n is divisible by 3
2. 2n is divisible by twice as many positive integers as n

Normally, one would have to say that n is odd, see
http://www.manhattangmat.com/forums/integer-t2061.html?hilit=tested%20on%20divisibility%20for%20positive%20numbers
but a question arose about 0 - "Can we consider n=0 for such division ?"

As background, I like to 'play with numbers'. So, I got to thinking, why there are just as many number between 0 and 3 as there is between 0 and 6 and 'obviously' there are twice as many numbers between 0 and 6 as there are between 0 and three. If I could prove that statement, along with a few others, it should be obvious to everyone that there must be twice as many divisors of 2n as there are divisors of n when n is zero.

First of all, how many divisors of 0 are there. We know there are an infinite number but what infinity? The same infinity as there are in the number of integers? Well, no actually there are many more. To get there we note that the sum of N items taken r at a time for r equal to 0, 1, 2, ..., N is 2^N. The "number of'" elements of the set of real numbers is the same as 2^N where N is the number of positive integers.

So we have that the number of factors of zero is the same as the number of real numbers. I know, this implies the number of factors of n is the same as the number of factors of 2n when n is zero but just bear with me.

Consider the function
tan[ (Ï€ / 2) * ( (x - a) / a)]
and lets just restrict our selves to positive a. When x is 0, the value is minus infinity [ tan[-(π / 2)] ]. When x is 2a, the value is plus infinity [ tan[ π / 2] ]. It is a strictly increasing function for x between 0 and 2a, so there is a one to one mapping between the interval of length 2a and between minus infinity and infinity. Well first let a = 6. This says that there are just as many points between x=0 and x=6 as there are between minus infinity and plus infinity. Thus there are the same number of elements in the set of factors of 2n when n is zero as there are the number of points (elements) between 0 and 6.

Now let a = 3/2. This shows that there are the same number of elements in the set of factors of n when n is zero as there are the number of points (elements) between 0 and 3.

Now obviously there are twice as many points in the interval between 0 and 6 as there are between 0 and 3. That being the case, there must be twice as many factors of 2n as there are of n.

Please note: I've been 'loose' in terminology and made one assumption unprovable in the generally accepted foundations of modern mathematics (assuming they are consistent). However, we are 'just having fun'.

[tim]

Fun post, John. I should probably warn people that this is way beyond GMAT material, just so no one panics. :) Anyway, yes the problem is flawed (is it one of ours?), not because it does not take 0 into account, but because the GMAT would never ask us to engage in the type of analysis required to test whether 0 is an acceptable value for n in this case..

To address your discussion about whether two infinities are the same, I’ll point out that the conventional definition of infinities indicates that two infinities are the same if they can be put into a one-to-one correspondence. In your example, if N is the set of integers, 2^N does indeed represent a different kind of infinity (note I’ll be offering a lot of mathematical facts here without proving or referencing them; they should be easy to look up though). However, you have not proven that the factors of 0 can be put into a correspondence with 2^N. Instead, they can be put into a correspondence with N. Regardless of whether the factors of 0 correspond to N or 2^N, doubling either of these infinities gives you the same infinity (not twice as much), so n cannot be 0 in this problem..

You actually did a very nice job of demonstrating that the number of points on any finite interval is equivalent to the number of points on the real number line. I’m curious then how you jumped from there to the contradictory conclusion that there are twice as many points on [0,3] as there are on [0,6]. Those two intervals in fact have the same number of points. I’ll leave it to you to come up with a function that maps one to the other.. :)