If $a,b,c$ are positive integers less than $10$, then $(10a + b)(10a + c) = 100a(a + 1) + bc$ if:
$\textbf{(A) }b+c=10$
$\qquad\textbf{(B) }b=c$
$\qquad\textbf{(C) }a+b=10$
$\qquad\textbf {(D) }a=b$
$\qquad\textbf{(E) }a+b+c=10$
Multiply out the LHS to get $100a^2+10ac+10ab+bc=100a(a+1)+bc$. Subtract $bc$ and factor to get $10a(10a+b+c)=10a(10a+10)$. Divide both sides by $10a$ and then subtract $10a$ to get $b+c=10$, giving an answer of $\boxed{A}$.