Tinem cont de faptul ca in total sunt 204 determeni,iar 204 este divizibil cu 2,3 si 4.(ne va ajuta asta mai tarziu)
[tex]S=1+3+3^2+_{\dots}+3^{203}\\
\text{Mai intai cu 3:}\\
S=(1+3)+3^2(1+3)+_{\dots}+ 3^{202}(1+3)\\
S=(1+3)(1+3^2+3^4+_{\dots}+3^{202})\\
S=4\cdot (1+3^2+3^4+_{\dots}+3^{202})\vdots\ 4\\
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[tex]\text{Apoi cu 13:}\\
S=(1+3+3^2)+3^3(1+3+3^2)+_{\dots}+3^{201}(1+3+3^2)\\
S=(1+3+3^2)(1+3^3+3^6+_{\dots}+3^{201})\\
S=13\cdot (1+3^3+3^6+_{\dots}+3^{201})\vdots\ 13\\
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\text{In cele din urma cu 10:}\\
S=(1+3+3^2+3^3)+3^4(1+3+3^2+3^3)+_{\dots}+3^{200}(1+3+3^2+3^3)\\
S=(1+3+3^2+3^3)(1+3^4+3^8+_{\dots}+3^{200})\\
S=40 \cdot (1+3^4+3^8+_{\dots}+3^{200})\\
S=10\cdot 4\cdot (1+3^4+3^8+_{\dots}+3^{200})\vdots 10[/tex]
