# Find all integers $n\in\mathbb{N}^*$ such that any integer $k$ satisfying $1\le k\le \sqrt{n}$ divides $n$

I've found following:

1, 2, 3, 4, 6, 8, 12, 24

and suspect that no integer larger than 24 satisfies the requirements.

How do I prove that or can you find a counterexample?

-
You don't need the floor function, because if $k$ is an integer, then $k \le E(x) \Leftrightarrow k \le x$. – TonyK Oct 11 '11 at 14:44
Uh ... yes, you are right. – Voldemort Oct 11 '11 at 14:47
It's weird for $k$ to be less than $1$ yet greater than $\sqrt n$... – J. M. Oct 11 '11 at 14:49
I think you mean $\le$ for both inequalities? – joriki Oct 11 '11 at 14:49
Why isn't 3 in the list? – Eric Naslund Oct 11 '11 at 14:54

look at the largest power of $2$, $3$, and $5$ that are under $\sqrt n$ : suppose $2^a, 3^b, 5^c \leq \sqrt n < 2^{a+1},3^{b+1},5^{c+1}$.

Then $2^a 3^b 5^c$ has to divide n, so you get the inequalities $n^{3/2}/30 = (\sqrt n/2)(\sqrt n/3)(\sqrt n/5)< 2^a 3^b 5^c \leq n$, and $\sqrt n < 30$. This means that such an $n$ has to be less than $900$.

You can add more primes into this and prove that $n \leq 173$. Then you can be more precise :

$2*3*5*7 > 173$, thus $\sqrt n < 7$, so $n<49$

$3*4*5 > 49$, thus $\sqrt n < 5$, so $n<25$

-
Did you just find this proof? That is brilliant! – Voldemort Oct 11 '11 at 15:38
Should it be "$3*4*5*7>173$"? – Voldemort Oct 11 '11 at 15:43