$\left \lfloor{\dfrac{m}{n}}\right \rfloor=a$ olsun. $$a\leq \dfrac{m}{n} <a+1$$ $$\Rightarrow an\leq m <an+n$$ $$an^2-a\leq mn-\left \lfloor{\dfrac{m}{n}}\right \rfloor <an^2+n^2-a$$ $$\Rightarrow a(n^2-1)\leq 51 <a(n^2-1)+n^2\Rightarrow n^2-1\leq 51\Rightarrow n\leq 7$$
$i)$ $n=7$ ise $48a\leq 51\Rightarrow a=1\Rightarrow 7m-1=51 \Rightarrow m=\dfrac{52}{7}$ Çelişki.
$ii)$ $n=6$ ise $35a\leq 51\Rightarrow a=1\Rightarrow 6m-1=51\Rightarrow m=\dfrac{26}{3}$ Çelişki.
$iii)$ $n=5$ ise $24a\leq 51<24a+25\Rightarrow a=2\Rightarrow 5m-2=51\Rightarrow m=\dfrac{53}{5}$ Çelişki.
$iv)$ $n=4$ ise $15a\leq 51<15a+16\Rightarrow a=3\Rightarrow 4m-3=51\Rightarrow m=\dfrac{27}{2}$ Çelişki.
$v)$ $n=3$ ise $8a\leq 51<8a+9\Rightarrow a=6\Rightarrow 3m-6=51\Rightarrow m=19\Rightarrow (m,n)=(19,3)$
$vi)$ $n=2$ ise $3a\leq 51<3a+4\Rightarrow a=16$ veya $a=17$
$via)$ $a=16$ ise $2m-16=51\Rightarrow m=\dfrac{67}{2}$ Çelişki.
$vib)$ $a=17$ ise $2m-17=51\Rightarrow m=34\Rightarrow (m,n)=(34,2)$
$vii)$ $n=1$ ise $mn-\lfloor{\dfrac{m}{n}}\rfloor=m-m=0$ Çelişki.
Çözümler $(m,n)=(19,3),(34,2)$'dir.
