If $a,b,c\in N$, such that $aaaaaa=b*c*bc*(b+b+c)*(c+c-b)(c-b)$, then find $a,b,c$.

Question

If $a,b,c\in N$, such that $aaaaaa=b*c*bc*(b+b+c)*(c+c-b)(c-b)$, then find $a,b,c$. Where $aaaaaa$ is a six digit no. with each digit equal to $a$ and $bc$ is a 2-digit no. with digits $b$ and $c$.

I can think of expressing $aaaaaa$ as $a\frac{10^6-1}{10-1}$, but how to proceed?

Answer 1

$aaaaaa=111111 \times a$. So lets factorise $111111=3 \times 7 \times 11 \times 13 \times 37$. Now check the factors on the right hand side ... $b=3$ , $ c=7$ ,$bc=37$ ,$(b+b+c)=13$,$(c+c-b)=11$, $(c-b)=4$. So $a= \color{red}{4}$.