Question

Determinați numărul tripletelor de cifre nenule(A,B,C), cu a<b<c,pentru care numărul
A=(a,b bară de deasupra +b,c bară deasupra +c,a bară deasupra este natural.
urgent ​

Answer 1

$\it\overline{a,b}+\overline{b,c}+\overline{c,a}=n|_{\cdot10} \Rightarrow \overline{ab}+\overline{bc}+\overline{ca}=10n \Rightarrow \\ \\ \Rightarrow 10a+b+10b+c+10c+a=10n \Rightarrow 11a+11b+11c=10n \Rightarrow \\ \\ \Rightarrow11(a+b+c)=10n \Rightarrow a+b+c\in M_{10}\ \ \ \ \ \ (1)\\ \\ a,\ b,\ c -cifre\ \stackrel{(1)}{\Longrightarrow} a+ b+ c\ \in\{10,\ \ 20\}\\ \\ \left.\begin{aligned}\ \it a+b+c=10\\ \\ \it a<b<c \end{aligned}\right\}\ \Rightarrow (a,\ b,\ c)\in\{(1,2,7),(1,3,6),(1,4,5),(2,3,5)\}$

$\it a+b+c=20 \Rightarrow (a,\ b,\ c)\in\{(5,\ 7,\ 8), (3,\ 8,\ 9), (4,7,9),\ (5, 6, 9)\}$