Explicație pas cu pas:
a)
[tex]{7}^{0} + {7}^{1} + {7}^{2} = 1 + 7 + 49 = \bf 57[/tex]
suma are 2004 de termeni, pe care îi putem grupa câte 3:
[tex]D = ({7}^{1} + {7}^{2} + {7}^{3}) + ({7}^{4} + {7}^{5} + {7}^{6}) + ... + ({7}^{2002} + {7}^{2003} + {7}^{2004}) = \\[/tex]
[tex]= 7 \cdot ({7}^{0} + {7}^{1} + {7}^{2}) + {7}^{4} \cdot ({7}^{0} + {7}^{1} + {7}^{2}) + ... + {7}^{2002} \cdot ({7}^{0} + {7}^{1} + {7}^{2}) \\ [/tex]
[tex]= ({7}^{0} + {7}^{1} + {7}^{2}) \cdot (7 + {7}^{4} + ... + {7}^{2002}) \\ [/tex]
[tex]= { \bf 57} \cdot (7 + {7}^{4} + ... + {7}^{2002} ) \ \red{ \bf \vdots \ 57} \\ [/tex]
=> D este divizibil cu 57
b)
[tex]{13}^{} + {13}^{2} = 13 + 169 = 182 = \bf 7 \cdot 26 \\ [/tex]
suma are 2012 de termeni, pe care îi putem grupa câte 2:
[tex]E = ({13}^{1} + {13}^{2}) + ({13}^{3} + ({13}^{4}) + ... + ({13}^{2011} + {13}^{2012}) = \\[/tex]
[tex]= ({13}^{1} + {13}^{2}) + {13}^{2} \cdot ({13}^{1} + {13}^{2}) + ... + {13}^{2010} \cdot ({13}^{1} + {13}^{2}) \\ [/tex]
[tex]= ({13}^{1} + {13}^{2}) \cdot (1 + {13}^{2} + ... + {13}^{2010}) \\ [/tex]
[tex]= 7 \cdot 26 \cdot (1 + {13}^{2} + ... + {13}^{2010})\ \red{ \bf \vdots \ 7} \\ [/tex]
=> E este divizibil cu 7
c)
[tex]1 + 2 + 3 + ... + 100 = \frac{100 \cdot 101}{2} = 5050 \\ [/tex]
[tex] \implies50500 \ \red{ \bf \vdots \ 5050} \\ [/tex]
d)
[tex]1 + 2 + 3 + ... + 20 = \frac{20 \cdot 21}{2} = 10 \cdot 21\ \red{ \bf \vdots \ 21} \\[/tex]
e)
[tex]{36}^{5} \cdot {6}^{5} = {6}^{x + 1} \iff{({6}^{2})}^{5} \cdot {6}^{5} = {6}^{x + 1} \\ {6}^{10} \cdot {6}^{5} = {6}^{x + 1} \iff {6}^{15} = {6}^{x + 1} \\ 15 = x + 1 \implies \red {\bf x = 14}[/tex]
