0
$\begingroup$

Find all the numbers $n$ $\in$ $\Bbb Z^+$ such that:

$$\left\lfloor\frac{n}{2}\right\rfloor \cdot \left\lfloor \frac{n}{3} \right\rfloor \cdot \left\lfloor \frac{n}{4} \right\rfloor = n^2$$

I never worked before with floor function so i'm not completely sure how to solve this. I think (only because $n$ $\in$ $\Bbb Z^+$), i can just multiply (skipping the floor function) and get the answer of $n=24$, but this is floor function so i don't know if there are more solutions.

Any hints?

$\endgroup$
  • 5
    $\begingroup$ It seems to me that the solution of $(\frac {n}{2} -1)(\frac {n}{3}-1)(\frac {n}{4} -1) = n^2$ should provide an upper bound for possible solutions, and this should give a manageable number of cases to check. $\endgroup$ – Robert Shore Feb 19 at 16:29
  • 1
    $\begingroup$ Yep, that's a good starting point. For all $n \ge 33$, we have $(\tfrac{n}{2}-1)(\tfrac{n}{3}-1)(\tfrac{n}{4}-1) > n^2$. Thus, we only need to check $1 \le n \le 32$. $\endgroup$ – JimmyK4542 Feb 19 at 16:31
  • 2
    $\begingroup$ Using the fact that $x-1\leq [x]<x$, we have $$ \frac{(n-2)(n-3)(n-4)}{24}\leq n^2<\frac{n^3}{24}$$ which implies $24\leq n$. $\endgroup$ – Qurultay Feb 19 at 16:33
  • $\begingroup$ And for $n \le 23$ we have $(n/2)(n/3)(n/4) < n^2$. So we don't need to check $n \le 23$. It suffices to check $24 \le n \le 32$. $\endgroup$ – Michael Lugo Feb 19 at 16:33
3
$\begingroup$

You already ofund a solution $n=24$. For $n<24$, we have $$ \left\lfloor\frac n2\right\rfloor \left\lfloor\frac n3\right\rfloor\left\lfloor\frac n4\right\rfloor\le \frac n{24}\cdot n^2<n^2.$$ For $n\ge 30$, we have $$\begin{align}\left\lfloor\frac n2\right\rfloor \left\lfloor\frac n3\right\rfloor\left\lfloor\frac n4\right\rfloor -n^2&\ge\frac{n-1}2\frac{n-2}3\frac{n-3}4-n^2\\&=\frac{n^3-30n^2+11n-6}{24}\\&\ge\frac{11n-6}{24}\\&>0.\end{align}$$ Hence we need at most check $25\le n\le 29$.

To simplify checks for these cases, note that $\left\lfloor\frac n2\right\rfloor$, $ \left\lfloor\frac n3\right\rfloor$, $\left\lfloor\frac n4\right\rfloor$ are three distinct integers $\ge 6$. As $29^2$ and $5^4$ cannot be written as product of three distinct integers $\>1$, we can exclude $n=29$ and $n=25$. For $3^6$, the only way to write it as product of three distinct integers is as $3\cdot 9\cdot 27$, but $3<6$, so we can exclude $n=27$ as well. For $n=26$, we see that only one of the three factors is a multiple of $13$, so this can be ruled out as well. Fially, for $n=28$, $\lfloor \frac n3\rfloor=9$, but $28^2$ is not a multiple of $3$.

$\endgroup$
  • $\begingroup$ (+1) I took too long working on my answer, that I did not see yours, which is essentially the same. $\endgroup$ – robjohn Feb 19 at 19:36
0
$\begingroup$

We have

$$ \left\lfloor\frac{n}{2}\right\rfloor \cdot \left\lfloor \frac{n}{3} \right\rfloor \cdot \left\lfloor \frac{n}{4} \right\rfloor - n^2 =0 $$

Note that: $$ x > \left\lfloor x \right\rfloor $$

Thus: $$ \frac{n}{2}\cdot \frac{n}{3}\cdot \frac{n}{4} - n^2 >\left\lfloor\frac{n}{2}\right\rfloor \cdot \left\lfloor \frac{n}{3} \right\rfloor \cdot \left\lfloor \frac{n}{4} \right\rfloor - n^2 = 0 $$

Thus:

$$ \frac{n^3}{24}-n^2= n^2\left( \frac{n}{24}-1\right) >0 $$

Hence we are looking for a number larger than 24.

At the same time: $$ \left\lfloor\frac{n}{2}\right\rfloor \cdot \left\lfloor \frac{n}{3} \right\rfloor \cdot \left\lfloor \frac{n}{4} \right\rfloor - n^2 > \left( \frac{n}{2}-1\right) \cdot\left( \frac{n}{3}-1\right) \cdot\left( \frac{n}{4}-1\right) -n^2 $$ Therefore: $$ 0>\left( \frac{n}{2}-1\right) \cdot\left( \frac{n}{3}-1\right) \cdot\left( \frac{n}{4}-1\right) -n^2 $$

This function has only one real root around 32.216 I'm lazy, sorry about that.

Hence we are looking for $n$ in the interval $[24,32]$. I'll check it in Python:

from math import floor
for n in range(24, 33):
    if (floor(n/2)*floor(n/3)*floor(n/4)-n*n) == 0:
        print(n)

And the only answer is 24.

$\endgroup$
0
$\begingroup$

A lengthy approach

As $[2,3,4]=12,$ let us try with all $12$ in-congruent residues $\pmod{12}$

If $n=12k$

$6k(4k)(3k)=(12k)^2\iff k=2$

If $n=12k+1,$

$6k(4k)(3k)=(12k+1)^2$ which is untenable as LHS is divisible by $6$

If $n=12k+2,(6k+1)(4k)(2k)=(12k+2)^2\iff6k+1=2k^2$ which is even

If $n=12k+3,(6k+1)(4k+1)2k=(12k+3)^2 which is odd unlike the LHS

If $n=12k+4,(12k+4)^2=(6k+2)(4k+1)2k\iff4(3k+1)=4k+1$ which is odd unlike LHS

and so on

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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