Question

A)BAratati ca nr 3+3²+3³+...+3^123 se divide cu 13
B)Rezolvati ecuatiile;
a)2x+xx=104
b)51x(cu bara de asupra)+5x1(cu bara de asupra)+x51(cu bara de asupra)=1395
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Answer 1

A)3¹+3²+3³+...+3¹²³=
=3(1+3¹+3²)+3⁴(1+3¹+3²)+...+3¹²¹(1+3¹+3²)=
=3*13+3⁴*13+...+3¹²¹*13=
=(3+3⁴+...+3¹²¹)*13   -> se divide cu 13

B)
a)$\overline{2x}+\overline{xx}=104$
20+x+10x+x=104
12x=84
x=7
Verificare: 
27+77=104

b)$\overline{51x}+\overline{5x1}+\overline{x51}=1395$
510+x+501+10x+100x+51=1395
111x=333
x=3
Verificare: 513+531+351=1395