Question

Aratati ca:
a)3 | a
a=31*$10 ^{n}$-25
b)3 | a
a=$10 ^{n+4}+10 ^{n+2} -41$
n∈N

Answer 1

a)[tex]a=31\cdot 10^n-31+6\\ a=31(10^n-1)+6\\ a=31\cdot \overline{999....9}+6\\ a=3(31\cdot \overline{33333...333}+2)\\ a\vdots3[/tex]
b)[tex]a=10^{n+4}+10^{n+2}-41\\ a=10^n(10^4+10^2)-41\\ a=10^n\cdot 10100-41\\ a=10^n\cdot 10100-10100+10100-41\\ a=10100(10^n-1)+10059\\ a=10100\cdot \overline{99999...99}+10059\\ a=3(10100\cdot \overline{333.33}+3353)\\ a\vdots3[/tex]