Question

Determinați numerele abc știind că a+b, b+c si c+a sunt direct proporționale cu numerele 3, 5 și 6.

Answer 1

$\{a, \ b, \ c, \} \ d.p. \ \{3, \ 5, \ 6 \}\\ \\ \Rightarrow \frac{a+b}{3}=\frac{b+c}{5}=\frac{c+a}{6}=\frac{a+b+b+c+c+a}{3+5+6}=\frac{a+b+c}{7}=k\\ \\ \Rightarrow a+b=3k\\ \\ \Rightarrow b+c=5k\\ \\ \Rightarrow a+c=6k\\ \\ \Rightarrow a+b+c=7k\\ \\ \\ \\ c=(a+b+c)-(a+b)=7k-3k=4k\\ \\ b=(a+b+c)-(a+c)=7k-6=1k\\ \\ a=(a+b+c)-(b+c)=7k-5k=2k$

$\Leftarrow \Rightarrow \frac{a}{2}=b=\frac{c}{4} \ | \cdot 4\\ \\ 2a=4b=c\\ \\ Obs: \ a, \ b \ si \ c \ sunt \ cifre! \\ \\ 1\leq c \leq 9\\ \\ \Rightarrow 1 \leq 2a \leq 9 \ |:2\\ \\ \Rightarrow \frac{1}{2}\leq a \leq \frac{9}{2}\\ \\ a\in \{1, \ 2, \ 3, \ 4, \ 5, \ 6, \ 7, \ 8, \ 9 \}\\ \\ \Rightarrow a \in \{1, \ 2, \ 3, \ 4 \}\\ \\ \\ \\ \Rightarrow 1\leq 4b \leq 9 \ |:4\\ \\ \Rightarrow \frac{1}{4}\leq b \leq \frac{9}{4}$

$b \in \{0, \ 1, \ 2, \ 3, \ 4, \ 5, \ 6, \ 7, \ 8, \ 9 \}\\ \\ \Rightarrow b \in \{1, \ 2 \}\\ \\ \\ Cazul \ I: \ b=1\\ \\ \Rightarrow 2a=4\cdot 1=c \Rightarrow a=\frac{4}{2}=2 \ si \ c=4\\ \\ Cazul \ II: \ b=2\\ \\ \Rightarrow 2a=4\cdot 2=c \Rightarrow a=\frac{8}{2}=2 \ si \ c=8\\ \\ \\ \\ Raspuns: \boxed{\overline{abc} \in \{ 214, \ 428 \}}$