Question

demonstrati ca 10 la puterea 9 |(1x2x3x...x40)

Answer 1

   
[tex]\displaystyle\\ 10^9~\Big|~(1\times2\times3\times \cdots \times 40) ~~~\text{unde }"\Big|" = "\text{divide}"\\\\ \text{ecvalent cu:}\\\\ (1\times2\times3\times \cdots \times 40)~\vdots ~ 10^9 ~~~\text{unde }"\vdots" = "\text{divizibil}"\\\\ \text{Pentru ca numarul }(1\times2\times3\times \cdots \times 40)~\text{sa fie divizivil ci } 10^9~ \text{, trebuie}\\ \text{ca acest numar sa se termine in cel putin 9 zerouri.} [/tex]


[tex]\displaystyle\\ \texttt{\bf Calculam numarul de zerouri in care se termina numarul: }\\ (1\times2\times3\times \cdots \times 40)\\\\ \text{Consideram ca descompunem in faxtori primi fiecare factor din numar.}\\\\ \text{Fiecare 0 (zero) de la sfarsitul produsului }(1\times2\times3\times \cdots \times 40) \\ \text{provine s\de la un produs dintre un 2 si un 5, acestia facand parte din}\\ \text{descompunerea in factori primi a factorilor de la 1 pana la 40}. [/tex]


[tex]\displaystile\\ \text{Gasim cate un 2 la fiecare numar par, dar cate un 5 il gasim mai rar.}\\ \text{Din acest motiv trebuie sa calculem numarul de aparitii a lui 5}\\ \text{in produsul: } (1\times2\times3\times \cdots \times 40).\\\\ ~~\:5 ........ 1 \text{ de } 5\\ 10 ........ 1 \text{ de } 5\\ 15 ........ 1 \text{ de } 5\\ 20 ........ 1 \text{ de } 5\\ 25 ........ 2 \text{ de } 5\\ 30 ........ 1 \text{ de } 5\\ 35 ........ 1 \text{ de } 5\\ 40 ........ 1 \text{ de } 5\\ [/tex]

[tex]\displaystyle\\ \text{In total 9 aparitii ale lui 5}\\ \Longrightarrow~~~\text{Produsul: } (1\times2\times3\times \cdots \times 40) ~\text{ se termina in 9 zerouri.}\\\\ \Longrightarrow~~~\boxed{\bf (1\times2\times3\times \cdots \times 40)~\vdots~10^9} \\\\ \Longrightarrow~~~\boxed{\bf 10^9~\Big|~(1\times2\times3\times \cdots \times 40)}[/tex]