Question

si a1a+a2a+a3a+......+a7a=mnp3base 8 calcule m+n+p+a

Answer 1

$\overline{a1a}+\overline{a2a}+\overline{a3a}+\cdots + \overline{a7a}=\overline{mnp3}_{(8)}\\ \\ (101a+10)+(101a+20)+(101a+30)+\cdots+(101a+70)=\overline{mnp3}_{(8)}\\ \\ 707a+(10+20+30+\cdots+70)=\overline{mnp3}_{(8)}\\ \\ 707a+280=\overline{mnp3}_{(8)}\\ \\ 707a+280\equiv 3 \mod 8\\ \\ 707a\equiv 3 \mod 8\hspace{1cm}\textit{(Pues 280 es m\'ultiplo de 8)}\\ \\ 3a\equiv 3 \mod 8\\ \\ \text{Candidatos a ser el valor de }a:\\ \\ \hspace*{4cm} a\in\{1,9\}\\ \\ \\ \text{Con }a=1$

$707a+280 = 707(1)+280\\ \\ 707a+280 = 987\\ \\ 707a+280 = 1733_{(8)}\\ \\ \\ \text{Con} a = 9\\ \\ 707a+280 = 707(9)+280\\ \\ 707a+280 = 6643\\ \\ 707a+280 = 14763_{(8)}\\ \\ \\ \text{Por ende }\\ \\ a=1\\ m=1\\ n=3\\ p=3\\ \\ \text{As\'i } \boxed{m+n+p+a=8}$