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I think you messed up the definition of "cute". IT does NOT mean devilishly hard.

how do you read n|N^2 ?? i seem to have forgotten

This is fun. Don't remove it, please. D:

N divides N^2

and n∤N ??

"does not divide"

Isnt N a number? How is it a set?

In this context, Set = let

Oh. Okay. Thanks!

wild guess 1??

Nothing is wild about this problem.

no ... i just find my logic too wild

I agree. There are certain things seeping into my brain.

@experimentx At least 15, man.

In the set of naturals, all numbers \(2^{i}\) with \(0\leq i \leq 9\) satisfy \(n < N\) and \(n \mid N\). Similarly, all numbers \(3^{i}\) with \(0\leq i \leq 9\). And likewise all values of \(f(i,j)=2^i3^j\) for all values of \(i\) and \(j\) between \(0\) and \(9\). This implies all \(n\) satisfying both \(n

Is it 84? Am I even close?

Argh. There are minor details that I messed up just now.

Close, very close :) But then I would appreciate a general and formal solution.

I am getting the feeling that \(\{n: n\mid N^2 \wedge n

Now, If it is some, say, \[3^{11} * 2^m\]

Actually, oops. \(\frac{k_1}{N}\) is not always an integer.

And similarly I did it for cases of 2 and .....
Obviously missed some.

The set \(\{n:(n

OOPS. I meant for the set to be:
\(\{n:(n

I think I can clarify my post:
If \(n \nmid N\), then \(\frac{N}{n} \neq k\) for some integer \(k\).
If \(\frac{N}{n} \neq k\) , then \(\frac{N}{n}=u\) for some \(u\) that is not an integer.
If \(n

Im getting 90 as an answer <.<

There is a little flaw in Limitless' argument. If n

I looked at the problem this way. If:\[n\mid N^2\]then n has to be of the form:\[n=2^a\cdot 3^b\]where:\[0\le a \le 20,0\le b\le 18\]But we dont want n to divide N, so a cant be less than 11 and b cant be less than 10 at the same time. Also, we dont want n to be bigger than N, so a cant be bigger than 9 and b cant be bigger than 8 at the same time. So what conditions on a and b are we looking for?
We want all n such that:\[n=2^a3^b\]such that only one of these two conditions are met: either:\[11\le a\le 20, 0\le b\le 8, n

Using the brute force method, @joemath314159 is correct. There are 90 possibilities.

Yeah, i could only brute force it too. Im not sure if there is a clever way to attempt this problem.

You could probably place some bounds on the sum of the exponents somehow.

90 is the right answer.

And there exists a very smart/easy solution. It is possible to generalize to any given N.

What is the elegant solution? :/