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PandaBear132
a fost răspuns

Sa se afle restul impartirii numarului : 1×2×3×4×...×2007×2008+2007 la 105

Răspuns :

Matepentrutoti
a=1×2×3×4×...×2007×2008+2007 =
=1×2×3×4×...×2007×2008+105x19+12=
=105x(1×2×3×4×...×2007×2008 +19)+12
Concluzie:
a:1×2×3×4×...×2007×2008+2007=105x(1×2×3×4×...×2007×2008 +19) rest 12.
 [tex](1*2*...*\fbox{105}*...*2008)\;\vdots\;105\;\\ \rightarrow\;restul\;impartirii\;va\fi\;determinat\;de \;\;2007:105=19\;\; (\it{rest=12})[/tex]

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