Question

Dau coroana si 40 puncte.Determinati numarul natural de 3 cifre distincte scris in baza 10 ,care este egal cu suma tuturor numerelor naturale de 2 cifre
distincte ce se pot forma cu cifrele sale.Vreau cu rezolvare cum ati facut

Answer 1

$Avem~deci~de~gasit~cifrele~distinctie~(in~baza~10)~a,b,c~astfel \\ \\ incat~ \overline{abc}= \overline{ab}+ \overline{ba}+ \overline{ac}+ \overline{ca}+ \overline{bc}+ \overline{cb}. \\ \\ Ecuatia~este~echivalenta~cu: \\ \\ 100a+10b+c=(10a+b)+(10b+a)+(10a+c)+(10c+a)+ \\ \\ +(10b+c)+(10c+b) \Leftrightarrow \\ \\ \Leftrightarrow 100a+10b+c=22a+22b+22c \Leftrightarrow \\ \\ \Leftrightarrow 100a-22a=22b+22c-10b-c \Leftrightarrow \\ \\ \Leftrightarrow 78a=12b+21c~~~~~(*)$

$Deoarece~b,c \leq9~(fiind~cifre~in~baza~10),~rezulta: \\ \\ 78a=12b+21c \leq 12 \cdot 9 +21 \cdot 9=297. \\ \\ Deci~78a \leq 297 \Rightarrow a \in \{1,2,3 \}.$

$Cazul~1:Daca~a=1,~relatia~(*)~devine:~78=12b+21c \Leftrightarrow \\ \\ \Leftrightarrow 2(39-6b)=21c.~De~aici~rezulta~ca~c~este~par.~ \\ \\ Totodata ~avem~21c \leq 78,~deci ~c \in \{0;2 \}. ~ \\ \\ c=0 \Rightarrow 78=12b,~imposibil!\\ \\c=2 \Rightarrow 78=12b+42,~si,~deci~b=3. \\ \\ Am~gasit~solutia~132. \\ \\ Cazul~2:Daca~a=2,~relatia~(*)~devine:~156=12b+21c \Leftrightarrow \\ \\ \Leftrightarrow4(39-3b)=21c.~De~aici~rezulta~ca~c~este~un~multiplu~de~4.$

$Insa ~21c \leq 156 \Rightarrow c \in \{0;4 \}. \\ \\ c=0 \Rightarrow 156=12b,~imposibil!~(caci~b \leq 9) \\ \\ c=4 \Rightarrow 156=12b+84 \Rightarrow b=6. \\ \\ Am~mai~gasit~solutia~264. \\ \\ Cazul~3:~Daca~a=3,~relatia~(*)~devine~234=12b+21c \Leftrightarrow \\ \\ \Leftrightarrow 6(39-2b)=21c. \Leftrightarrow 2(39-2b)=7c.~Deci~c~este~un~multiplu \\ \\ de~3~par;~si~cum ~c \neq a,~rezulta ~c \in \{0;6\}. \\ \\ c=0 \Rightarrow 234=12b,~imposibil! \\ \\ c=6 \Rightarrow 234=12b +126 \Rightarrow b=9.$

$Am~mai~gasit~deci~solutia~396. \\ \\ SOLUTIE:~132,~264,~396.$