Question

Demonstrați că 10 la a noua divide cu 1 ori 2 ori 3 ori puncte puncte ori 40​

Answer 1

 

$!0^9~|~1\times2\times3\times...\times40\\\\\text{Trebuie sa aratam ca produsul numerelor de la 1 la 40 este divizibil cu}~ 10^9\\\\\text{Asta inseamna ca produsul se termina in cel putin 9 zerouri.}\\\\\text{Calculam numarul de zerouri in care se termina produsul.}\\\\\text{Explicatii:}\\\\\text{Fiecare zero de la sfarsitul produsului apare datorita produsului:}\\ 2\times5=10\\\\\text{In produs sunt mai multe numere pare decat multipli lui 5.}$

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$\displaystyle\\\text{Rezulta ca e suficient sa aflam numarul de cinciuri din produs.}\\\\\text{\bf~Metoda 1:}\\\\\bf\\~~~5~~~1~cinci\\10~~~1~cinci\\15~~~1~cinci\\20~~~1~cinci\\25~~~2~cinciuri\\30~~~1~cinci\\35~~~1~cinci\\40~~~1~cinci\\Total~9~cinciuri\\\text{\bf Rezulta ca produsul se termina in 9 zerouri.}\\\implies~\boxed{\bf~(1\times2\times3\times...\times40)~\vdots~10^9}$

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$\displaystyle\bf\\Metoda~2:\\n=numarul~de~cinciuri.\\\\n=\left[\frac{40}{5}\right]+\left[\frac{40}{5^2}\right]=8+1=9~cinciuri\\\text{\bf Nu am folosit }5^3~deoarece~5^3>40\\\\Parantezele~\Big[~\Big]~inseamna~partea~intreaga.\\n=9~cinciuri\\\\\text{\bf Rezulta ca produsul se termina in 9 zerouri.}\\\implies~\boxed{\bf~(1\times2\times3\times...\times40)~\vdots~10^9}$