Exercițiul 702

E.702. Determinați numerele de forma abcdef‾,\overline{abcdef}, știind că {a,b,c,d,e,f}={1,2,3,4,5,6}, abcdef‾ ⋮ 6, abcde‾ ⋮ 5, abcd‾ ⋮ 4, abc‾ ⋮ 3, ab‾ ⋮ 2.\{a,b,c,d,e,f\} = \{1,2,3,4,5,6\},~ \overline{abcdef} ~\vdots~6, ~ \overline{abcde} ~\vdots~5,~ \overline{abcd} ~\vdots~4, ~ \overline{abc} ~\vdots~3, ~ \overline{ab} ~\vdots~2.

Olimpiadă, etapa locală, Neamț, 2020
Soluție
Soluție:

{a,b,c,d,e,f}={1,2,3,4,5,6},\{a,b,c,d,e,f\} = \{1,2,3,4,5,6\}, deci cifrele nu se repetă.

  • abcde‾ ⋮ 5⇒e=5\overline{abcde} ~\vdots~ 5 \Rightarrow \boxed{e=5} și {a,b,c,d,f}={1,2,3,4,6};\{a,b,c,d,f\} = \{1,2,3,4,6\};
  • abcdef‾ ⋮ 6, abcd‾ ⋮ 4, ab‾ ⋮ 2⇒{b,d,f}={2,4,6}⇒{a,c}={1,3};\overline{abcdef} ~\vdots~6, ~ \overline{abcd} ~\vdots~4, ~ \overline{ab} ~\vdots~2 \Rightarrow \boxed{\{b,d,f\} = \{2,4,6\}} \Rightarrow \boxed{\{a,c\} = \{1,3\}};
  • abc‾ ⋮ 3⇒(a+b+c) ⋮ 3⇒(4+b) ⋮ 3⇒b=2⇒{d,f}={4,6};\overline{abc} ~\vdots~3 \Rightarrow (a+b+c) ~\vdots~ 3 \Rightarrow (4+b) ~\vdots~ 3 \Rightarrow \boxed{b=2} \Rightarrow \boxed{\{d,f\} = \{4,6\}};
  • abcd‾ ⋮ 4⇒cd‾ ⋮ 4⇒cd‾∈{16,36}⇒d=6⇒f=4;\overline{abcd} ~\vdots~4 \Rightarrow \overline{cd} ~\vdots~4 \Rightarrow \overline{cd} \in \{16,36\} \Rightarrow \boxed{d=6} \Rightarrow \boxed{f=4};
    • dacă c=1⇒a=3→abcdef‾=321654;\boxed{c=1} \Rightarrow \boxed{a=3} \rightarrow \boxed{\overline{abcdef} = 321654};
    • dacă c=3⇒a=1→abcdef‾=123654.\boxed{c=3} \Rightarrow \boxed{a=1} \rightarrow \boxed{\overline{abcdef} = 123654}.