Notăm numerele cerute cu a și b.
[tex]\it a:b=3\ rest\ r \Rightarrow \begin{cases} \it a=3b+r\ \ \ \ \ (1)\\ \\ \it r<b\ \ \ \ \ (2)\end{cases}\\ \\ \\ a+b=36 \stackrel{(1)}{\Longrightarrow} 3b+r+b=36 \Rightarrow 4b+r=36 \Rightarrow r = 36-4b \Rightarrow \\ \\ \Rightarrow r=4k,(k\in\mathbb{N}) \Rightarrow r\in\{0,\ 4,\ 8, ...\} \\ \\ r=36-4b\Rightarrow 4b=36-r \Rightarrow b=(36-r):4\ \ \ \ (3)[/tex]
[tex]\it r=0 \stackrel{(3)}{\Longrightarrow}\ b=(36-0):4=36:4=9 \stackrel{(1)}{\Longrightarrow} a=3\cdot9+0 \Rightarrow a=27\\ \\ \\ r=4 \stackrel{(3)}{\Longrightarrow}\ b=(36-4):4=32:4=8 \stackrel{(1)}{\Longrightarrow} a=3\cdot8+4 \Rightarrow a = 28\\ \\ r=8 \stackrel{(3)}{\Longrightarrow}\ b=(36-8):4=28:4=7 \stackrel{(2)}{\Longrightarrow}\ nu\ convine,\ ( r>b)[/tex]
Problema admite două soluții :
I) a=27, b=9;
II) a=28, b=8.
