od ms.srki » Subota, 09. Maj 2015, 09:13
POZIV NA MATEMATIČKI MEGDAN ZA Daniela ( pošto ima najviše matematičkog znanja iz sadašnje matematike , može i neko drugi koji zna odgovore ) , kako bi on to rešio , imamo [inlmath]a=\{0,4\}\cup\{6,9\}\cup\{11,13\}[/inlmath] , [inlmath]b=\{0,3\}\cup\{5,10\}[/inlmath] , treba da dobijemo [inlmath]c_n[/inlmath]
[inlmath]c_1=(1,1,1,2). moja. notacija. i. resenje. 4\underline23\underline22{+_1^{\underline0}}3 \underline25=(1,1,1,2)[/inlmath]
[inlmath]c_2=(1,4) --4\underline23\underline22{+_1^{\underline1}}3 \underline25=(1,4)[/inlmath]
[inlmath]c_3=(2,1,1,2,1)--4\underline23\underline22{+_1^{\underline2}}3 \underline25=(2,1,1,2,1)[/inlmath]
[inlmath]c_4=(3,4,2)--4\underline23\underline22{+_1^{\underline3}}3 \underline25=(3,4,2)[/inlmath]
[inlmath]c_5=(6,4,1)--4\underline23\underline22{+_1^{\underline4}}3 \underline25=(6,4,1)[/inlmath]
[inlmath]c_6=(4,1,1,1,2)--4\underline23\underline22{+_1^{\underline5}}3 \underline25=(4,1,1,1,2)[/inlmath]
[inlmath]c_7=(4,3)--4\underline23\underline22{+_1^{\underline6}}3 \underline25=(4,3)[/inlmath]
[inlmath]c_8=(4,1,1,1,4)--4\underline23\underline22{+_1^{\underline7}}3 \underline25=(4,1,1,1,4)[/inlmath]
[inlmath]c_9=(4,2,9)--4\underline23\underline22{+_1^{\underline8}}3 \underline25=(4,2,9)[/inlmath]
[inlmath]c_{10}=(4,5,5)--4\underline23\underline22{+_1^{\underline9}}3 \underline25=(4,5,5)[/inlmath]
[inlmath]c_{11}=(4,3,1,5)--4\underline23\underline22{+_1^{\underline{10}}}3 \underline25=(4,3,1,5)[/inlmath]
[inlmath]c_{12}=(4,3,1,5)--4\underline23\underline22{+_1^{\underline{11}}}3 \underline25=(4,3,1,5)[/inlmath]
[inlmath]c_{13}=(4,3,2,5)--4\underline23\underline22{+_1^{\underline{12}}}3 \underline25=(4,3,2,5)[/inlmath]
[inlmath]c_{14}=(4,3,3,5)--4\underline23\underline22{+_1^{\underline{13}}}3 \underline25=(4,3,3,5)[/inlmath]