Question

Determinati cel mai mare numar natural ,,n'' astfel incat 2 la ,,n'' divizibil ( 2 x 4 x 6 x......x50 dau coronitaa!!! Urgent!)

Answer 1

    
[tex]Cautam ~~~n_{maxim} ~pentru ~care~~2^n~|~( 2 \times 4 \times 6 \times......\times 50) \\ adica ~~( 2 \times 4 \times 6 \times......\times 50) ~\vdots~2^n \\ \\ 2 \times 4 \times 6 \times......\times 50 = \\ = 2 \cdot 1 \times 2 \cdot 2 \times 2 \cdot 3 \times......\times 2 \cdot 25 = \\ = 2^{25}(1\times 2\times 3 \times ......\times 25)= \\ =2^{25}(2\times 4\times 6 \times ......\times 24)(1\times 3\times 5\times ......\times 25)= [/tex]

[tex]=2^{25}(2 \cdot 1 \times 2 \cdot 2 \times 2 \cdot 3 \times ......\times 2 \cdot 12)(1\times 3\times 5\times ......\times 25)= \\ = 2^{25} \cdot 2^{12}(1 \times 2 \times 3 \times ... \times12)(1\times 3\times 5\times ...\times 25)= \\ = 2^{25} 2^{12}(2 \times 4 \times 6 \times ... \times12)(1 \times 3 \times 5 \times ... \times11)(1\times 3\times 5\times ...\times 25) \\ Eliminam~factorii~formati~din~numere~prime: \\ =2^{25} \cdot 2^{12}(2\cdot1 \times 2\cdot2 \times 2\cdot3 \times ... \times2\cdot6)= [/tex]

[tex]=2^{25} \cdot 2^{12}(2\cdot1 \times 2\cdot2 \times 2\cdot3 \times ... \times2\cdot6)= \\ =2^{25} \cdot 2^{12} \cdot 2^{6}(1\times2\times3\times...\times6)= \\ =2^{25} \cdot 2^{12} \cdot 2^{6}(2\times4\times6)(1\times3\times5) = (Eliminam~ultima~paranteza) \\ = 2^{25} \cdot 2^{12} \cdot 2^{6}(2\times4\times6) = 2^{25} \cdot 2^{12} \cdot 2^{6} \cdot 2^4 = \\ = 2^{25+12+6+4} = \boxed{2^{47}}~(inmultit ~cu ~multe~ numere~ inpare) \\ =\ \textgreater \ ~~ n_{maxim} = \boxed{47}[/tex]