if n>11 LHS>RHS。so check n<=10

parmenides51 wrote: Find all postitive integers n such that $$\left\lfloor \frac{n}{2} \right\rfloor \cdot \left\lfloor \frac{n}{3} \right\rfloor \cdot \left\lfloor \frac{n}{4} \right\rfloor=n^2$$where $\lfloor x \rfloor$ represents the largest integer less than the real number $x$. $\left\lfloor \frac{n}{2} \right\rfloor > \frac{n}{2} -1 \ge \frac{n}{8} \hspace{0.25cm} \forall n \geq 3$ $\left\lfloor \frac{n}{3} \right\rfloor > \frac{n}{3} - 1 \ge \frac{n}{8} \hspace{0.25cm} \forall n \geq 5$ $\left\lfloor \frac{n}{4} \right\rfloor > \frac{n}{4} - 1 \ge \frac{n}{8} \hspace{0.25cm} \forall n \geq 8$ so $$\left\lfloor \frac{n}{2} \right\rfloor \cdot \left\lfloor \frac{n}{3} \right\rfloor \cdot \left\lfloor \frac{n}{4} \right\rfloor=n^2 \ge \frac{n^3}{8}$$$\iff 8 \ge n$ Therefore, check $n=1,2,3,4,5,6,7,8$

masterhuang wrote: if n>11 LHS>RHS。so check n<=10 straight wrote: Therefore, check $n=1,2,3,4,5,6,7,8$ What about $n=24$ ?

pco wrote: masterhuang wrote: if n>11 LHS>RHS。so check n<=10 straight wrote: Therefore, check $n=1,2,3,4,5,6,7,8$ What about $n=24$ ? Oh，if n>29 LHS>RHS,I was wrong just now.

$n \geq 24$ by $[x] \leq x$ $[\frac{n}{2} ] >\frac{n}{2}-1 \geq \frac{11n}{24}$ $[\frac{n}{3}] > \frac{n}{3}-1 \geq \frac{7n}{24}$ $[\frac{n}{4}] >\frac{n}{4}-1 \geq \frac{5n}{24}$ , $ n \geq 24$ $n<35$ , $n\leq 34$ $n = 24+r , r \in [0,1,....,10]$ $r(r+19)<114$ $0\leq r \leq 4$ Bash $r=0,r=1,r=2,r=3,r=4$ $n=24,n=25,n=26,n=27,n=28$ $n=24$ True $n=25 , 12*8*6=25^2$ False since lhs doesnt contain 5 $n=26,13*8*6 = 26^2$ False since lhs doesnt contain $13^2$ $n=27, 13*9* 6 = 27^2$ RHS doesnt contain 13 $n=28 , 14*9*7 = 28^2$ RHS doesnt contin 9 thus n=24 is the only sols

parmenides51 wrote: Find all postitive integers n such that $$\left\lfloor \frac{n}{2} \right\rfloor \cdot \left\lfloor \frac{n}{3} \right\rfloor \cdot \left\lfloor \frac{n}{4} \right\rfloor=n^2$$where $\lfloor x \rfloor$ represents the largest integer less than the real number $x$. We claim the answer is $24.$ It clearly works. Clearly, the function $LHS-RHS$ is at most $\frac{n^3}{24}-n^2,$ so if $n<24$ it must be negative. Thus, $n\geq 24.$ Now consider $f(x)=\frac{x-1}{2}\frac{x-2}{3}\frac{x-3}{4}-x^2.$ This is the lower bound for the $LHS-RHS.$ This is $\frac{x^3-30x^2+11x-6}{24}.$ Note now that $f(30)>0.$ $f(x)$ has derivative $\frac{3x^2-60x+11}{24}.$ If $x>30,$ we note that this is $\frac{3x(x-20)+11}{24}>0.$ Thus, $x$ is at most $29.$ Note that if $n=2k+1,$ then $k$ must divide $(2k+1)^2,$ then $k$ is $1.$ Thus, odd $n$ fails. We have $26$ and $28$ left. One can easily check they fail. Edit: sniped

pco wrote: masterhuang wrote: if n>11 LHS>RHS。so check n<=10 straight wrote: Therefore, check $n=1,2,3,4,5,6,7,8$ What about $n=24$ ? Oh bruh, I wrote that $8^3 = 8$. I feel smart

Ugly method but it work: we have $n=24k+r$ with $r\in \left\{ 0,\pm 1,\pm 2,\pm 3,...,\pm 11 \right\}$, and casework.