Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 6 down vote accepted

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 $n<25$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.