Explicație pas cu pas:
[tex]AB = {4}^{10} \times {3}^{20} = {2}^{20} \times {3}^{20} = {6}^{20}[/tex]
a)
[tex]AA_{1} = 5A_{1}B \\ AB = AA_{1} + A_{1}B = 6A_{1}B \\ A_{1}B = \frac{AB}{6} = \frac{ {6}^{20} }{6} \implies \bf A_{1}B = {6}^{19} [/tex]
[tex]A_{1}A_{2} = 5A_{2}B \\ A_{1}B = A_{1}A_{2} + A_{2}B = 6A_{2}B \\ A_{2}B = \frac{A_{1}B}{6} = \frac{ {6}^{19} }{6} \implies \bf A_{2}B = {6}^{18} \\ \implies \red{\bf A_{1}A_{2} = 5 \cdot {6}^{18}}[/tex]
b)
[tex]A_{1}B = \frac{1}{6} \cdot AB \\ A_{2}B = \frac{1}{6} \cdot A_{1}B = \frac{1}{6} \cdot \frac{1}{6} \cdot AB = {\Big( \frac{1}{6} \Big)}^{2} \cdot AB \\ A_{3}B = \frac{1}{6} \cdot A_{2}B = {\Big( \frac{1}{6} \Big)}^{3} \cdot AB \\ ... \\ A_{10}B = {\Big( \frac{1}{6} \Big)}^{10} \cdot AB = \frac{ {6}^{20} }{ {6}^{10} }\implies \red{ \bf A_{10}B = {6}^{10}} [/tex]
